Electric Fields. Chapter Outline Properties of Electric Charges 23.2 Insulators and Conductors 23.3 Coulomb s Law 23.4 The Electric Field

Transcription

1 P U Z Z L E R Soft contct lenses re comfortble to wer becuse they ttrct the proteins in the werer s ters, incorporting the complex molecules right into the lenses. They become, in sense, prt of the werer. Some types of mkeup exploit this sme ttrctive force to dhere to the skin. Wht is the nture of this force? (Chrles D. Winters) c h p t e r Electric Fields Chpter Outline 23.1 Properties of Electric Chrges 23.2 Insultors nd Conductors 23.3 Coulomb s Lw 23.4 The Electric Field 23.5 Electric Field of Continuous Chrge Distribution 23.6 Electric Field Lines 23.7 Motion of Chrged Prticles in Uniform Electric Field 708

2 T he electromgnetic force between chrged prticles is one of the fundmentl forces of nture. We begin this chpter by describing some of the bsic properties of electric forces. We then discuss Coulomb s lw, which is the fundmentl lw governing the force between ny two chrged prticles. Next, we introduce the concept of n electric field ssocited with chrge distribution nd describe its effect on other chrged prticles. We then show how to use Coulomb s lw to clculte the electric field for given chrge distribution. We conclude the chpter with discussion of the motion of chrged prticle in uniform electric field Properties of Electric Chrges PROPERTIES OF ELECTRIC CHARGES A number of simple experiments demonstrte the existence of electric forces nd chrges. For exmple, fter running comb through your hir on dry dy, you will find tht the comb ttrcts bits of pper. The ttrctive force is often strong enough to suspend the pper. The sme effect occurs when mterils such s glss or rubber re rubbed with silk or fur. Another simple experiment is to rub n inflted blloon with wool. The blloon then dheres to wll, often for hours. When mterils behve in this wy, they re sid to be electrified, or to hve become electriclly chrged. You cn esily electrify your body by vigorously rubbing your shoes on wool rug. The electric chrge on your body cn be felt nd removed by lightly touching (nd strtling) friend. Under the right conditions, you will see sprk when you touch, nd both of you will feel slight tingle. (Experiments such s these work best on dry dy becuse n excessive mount of moisture in the ir cn cuse ny chrge you build up to lek from your body to the Erth.) In series of simple experiments, it is found tht there re two kinds of electric chrges, which were given the nmes positive nd negtive by Benjmin Frnklin ( ). To verify tht this is true, consider hrd rubber rod tht hs been rubbed with fur nd then suspended by nonmetllic thred, s shown in Figure When glss rod tht hs been rubbed with silk is brought ner the rubber rod, the two ttrct ech other (Fig. 23.1). On the other hnd, if two chrged rubber rods (or two chrged glss rods) re brought ner ech other, s shown in Figure 23.1b, the two repel ech other. This observtion shows tht the rubber nd glss re in two different sttes of electrifiction. On the bsis of these observtions, we conclude tht like chrges repel one nother nd unlike chrges ttrct one nother. Using the convention suggested by Frnklin, the electric chrge on the glss rod is clled positive nd tht on the rubber rod is clled negtive. Therefore, ny chrged object ttrcted to chrged rubber rod (or repelled by chrged glss rod) must hve positive chrge, nd ny chrged object repelled by chrged rubber rod (or ttrcted to chrged glss rod) must hve negtive chrge. Attrctive electric forces re responsible for the behvior of wide vriety of commercil products. For exmple, the plstic in mny contct lenses, etfilcon, is mde up of molecules tht electriclly ttrct the protein molecules in humn ters. These protein molecules re bsorbed nd held by the plstic so tht the lens ends up being primrily composed of the werer s ters. Becuse of this, the werer s eye does not tret the lens s foreign object, nd it cn be worn comfortbly. Mny cosmetics lso tke dvntge of electric forces by incorporting mterils tht re electriclly ttrcted to skin or hir, cusing the pigments or other chemicls to sty put once they re pplied. QuickLb Rub n inflted blloon ginst your hir nd then hold the blloon ner thin strem of wter running from fucet. Wht hppens? (A rubbed plstic pen or comb will lso work.)

3 710 CHAPTER 23 Electric Fields Rubber Rubber Figure 23.1 F F Glss () F F Rubber () A negtively chrged rubber rod suspended by thred is ttrcted to positively chrged glss rod. (b) A negtively chrged rubber rod is repelled by nother negtively chrged rubber rod. (b) Chrge is conserved Another importnt spect of Frnklin s model of electricity is the impliction tht electric chrge is lwys conserved. Tht is, when one object is rubbed ginst nother, chrge is not creted in the process. The electrified stte is due to trnsfer of chrge from one object to the other. One object gins some mount of negtive chrge while the other gins n equl mount of positive chrge. For exmple, when glss rod is rubbed with silk, the silk obtins negtive chrge tht is equl in mgnitude to the positive chrge on the glss rod. We now know from our understnding of tomic structure tht negtively chrged electrons re trnsferred from the glss to the silk in the rubbing process. Similrly, when rubber is rubbed with fur, electrons re trnsferred from the fur to the rubber, giving the rubber net negtive chrge nd the fur net positive chrge. This process is consistent with the fct tht neutrl, unchrged mtter contins s mny positive chrges (protons within tomic nuclei) s negtive chrges (electrons). Figure 23.2 Rubbing blloon ginst your hir on dry dy cuses the blloon nd your hir to become chrged. Chrge is quntized Quick Quiz 23.1 If you rub n inflted blloon ginst your hir, the two mterils ttrct ech other, s shown in Figure Is the mount of chrge present in the blloon nd your hir fter rubbing () less thn, (b) the sme s, or (c) more thn the mount of chrge present before rubbing? In 1909, Robert Millikn ( ) discovered tht electric chrge lwys occurs s some integrl multiple of fundmentl mount of chrge e. In modern terms, the electric chrge q is sid to be quntized, where q is the stndrd symbol used for chrge. Tht is, electric chrge exists s discrete pckets, nd we cn write q Ne, where N is some integer. Other experiments in the sme period showed tht the electron hs chrge e nd the proton hs chrge of equl mgnitude but opposite sign e. Some prticles, such s the neutron, hve no chrge. A neutrl tom must contin s mny protons s electrons. Becuse chrge is conserved quntity, the net chrge in closed region remins the sme. If chrged prticles re creted in some process, they re lwys creted in pirs whose members hve equl-mgnitude chrges of opposite sign.

4 23.2 Insultors nd Conductors 711 From our discussion thus fr, we conclude tht electric chrge hs the following importnt properties: Two kinds of chrges occur in nture, with the property tht unlike chrges ttrct one nother nd like chrges repel one nother. Chrge is conserved. Chrge is quntized. Properties of electric chrge INSULATORS AND CONDUCTORS It is convenient to clssify substnces in terms of their bility to conduct electric chrge: Electricl conductors re mterils in which electric chrges move freely, wheres electricl insultors re mterils in which electric chrges cnnot move freely. Mterils such s glss, rubber, nd wood fll into the ctegory of electricl insultors. When such mterils re chrged by rubbing, only the re rubbed becomes chrged, nd the chrge is unble to move to other regions of the mteril. In contrst, mterils such s copper, luminum, nd silver re good electricl conductors. When such mterils re chrged in some smll region, the chrge redily distributes itself over the entire surfce of the mteril. If you hold copper rod in your hnd nd rub it with wool or fur, it will not ttrct smll piece of pper. This might suggest tht metl cnnot be chrged. However, if you ttch wooden hndle to the rod nd then hold it by tht hndle s you rub the rod, the rod will remin chrged nd ttrct the piece of pper. The explntion for this is s follows: Without the insulting wood, the electric chrges produced by rubbing redily move from the copper through your body nd into the Erth. The insulting wooden hndle prevents the flow of chrge into your hnd. Semiconductors re third clss of mterils, nd their electricl properties re somewhere between those of insultors nd those of conductors. Silicon nd germnium re well-known exmples of semiconductors commonly used in the fbriction of vriety of electronic devices, such s trnsistors nd light-emitting diodes. The electricl properties of semiconductors cn be chnged over mny orders of mgnitude by the ddition of controlled mounts of certin toms to the mterils. When conductor is connected to the Erth by mens of conducting wire or pipe, it is sid to be grounded. The Erth cn then be considered n infinite sink to which electric chrges cn esily migrte. With this in mind, we cn understnd how to chrge conductor by process known s induction. To understnd induction, consider neutrl (unchrged) conducting sphere insulted from ground, s shown in Figure When negtively chrged rubber rod is brought ner the sphere, the region of the sphere nerest the rod obtins n excess of positive chrge while the region frthest from the rod obtins n equl excess of negtive chrge, s shown in Figure 23.3b. (Tht is, electrons in the region nerest the rod migrte to the opposite side of the sphere. This occurs even if the rod never ctully touches the sphere.) If the sme experiment is performed with conducting wire connected from the sphere to ground (Fig. 23.3c), some of the electrons in the conductor re so strongly repelled by the presence of Metls re good conductors Chrging by induction

5 712 CHAPTER 23 Electric Fields () (b) (c) (d) Figure 23.3 Chrging metllic object by induction (tht is, the two objects never touch ech other). () A neutrl metllic sphere, with equl numbers of positive nd negtive chrges. (b) The chrge on the neutrl sphere is redistributed when chrged rubber rod is plced ner the sphere. (c) When the sphere is grounded, some of its electrons leve through the ground wire. (d) When the ground connection is removed, the sphere hs excess positive chrge tht is nonuniformly distributed. (e) When the rod is removed, the excess positive chrge becomes uniformly distributed over the surfce of the sphere. (e)

6 23.3 Coulomb s Lw 713 Chrged object Insultor Induced chrges QuickLb Ter some pper into very smll pieces. Comb your hir nd then bring the comb close to the pper pieces. Notice tht they re ccelerted towrd the comb. How does the mgnitude of the electric force compre with the mgnitude of the grvittionl force exerted on the pper? Keep wtching nd you might see few pieces jump wy from the comb. They don t just fll wy; they re repelled. Wht cuses this? Figure 23.4 () (b) () The chrged object on the left induces chrges on the surfce of n insultor. (b) A chrged comb ttrcts bits of pper becuse chrges re displced in the pper the negtive chrge in the rod tht they move out of the sphere through the ground wire nd into the Erth. If the wire to ground is then removed (Fig. 23.3d), the conducting sphere contins n excess of induced positive chrge. When the rubber rod is removed from the vicinity of the sphere (Fig. 23.3e), this induced positive chrge remins on the ungrounded sphere. Note tht the chrge remining on the sphere is uniformly distributed over its surfce becuse of the repulsive forces mong the like chrges. Also note tht the rubber rod loses none of its negtive chrge during this process. Chrging n object by induction requires no contct with the body inducing the chrge. This is in contrst to chrging n object by rubbing (tht is, by conduction), which does require contct between the two objects. A process similr to induction in conductors tkes plce in insultors. In most neutrl molecules, the center of positive chrge coincides with the center of negtive chrge. However, in the presence of chrged object, these centers inside ech molecule in n insultor my shift slightly, resulting in more positive chrge on one side of the molecule thn on the other. This relignment of chrge within individul molecules produces n induced chrge on the surfce of the insultor, s shown in Figure Knowing bout induction in insultors, you should be ble to explin why comb tht hs been rubbed through hir ttrcts bits of electriclly neutrl pper nd why blloon tht hs been rubbed ginst your clothing is ble to stick to n electriclly neutrl wll. Quick Quiz 23.2 Object A is ttrcted to object B. If object B is known to be positively chrged, wht cn we sy bout object A? () It is positively chrged. (b) It is negtively chrged. (c) It is electriclly neutrl. (d) Not enough informtion to nswer COULOMB S LAW Chrles Coulomb ( ) mesured the mgnitudes of the electric forces between chrged objects using the torsion blnce, which he invented (Fig. 23.5). Chrles Coulomb ( ) Coulomb's mjor contribution to science ws in the field of electrosttics nd mgnetism. During his lifetime, he lso investigted the strengths of mterils nd determined the forces tht ffect objects on bems, thereby contributing to the field of structurl mechnics. In the field of ergonomics, his reserch provided fundmentl understnding of the wys in which people nd nimls cn best do work. (Photo courtesy of AIP Niels Bohr Librry/E. Scott Brr Collection)

7 714 CHAPTER 23 Electric Fields B A Figure 23.5 Coulomb constnt Suspension hed Fiber Coulomb s torsion blnce, used to estblish the inverse-squre lw for the electric force between two chrges. Chrge on n electron or proton Coulomb confirmed tht the electric force between two smll chrged spheres is proportionl to the inverse squre of their seprtion distnce r tht is, F e 1/r 2. The operting principle of the torsion blnce is the sme s tht of the pprtus used by Cvendish to mesure the grvittionl constnt (see Section 14.2), with the electriclly neutrl spheres replced by chrged ones. The electric force between chrged spheres A nd B in Figure 23.5 cuses the spheres to either ttrct or repel ech other, nd the resulting motion cuses the suspended fiber to twist. Becuse the restoring torque of the twisted fiber is proportionl to the ngle through which the fiber rottes, mesurement of this ngle provides quntittive mesure of the electric force of ttrction or repulsion. Once the spheres re chrged by rubbing, the electric force between them is very lrge compred with the grvittionl ttrction, nd so the grvittionl force cn be neglected. Coulomb s experiments showed tht the electric force between two sttionry chrged prticles is inversely proportionl to the squre of the seprtion r between the prticles nd directed long the line joining them; is proportionl to the product of the chrges q 1 nd q 2 on the two prticles; is ttrctive if the chrges re of opposite sign nd repulsive if the chrges hve the sme sign. From these observtions, we cn express Coulomb s lw s n eqution giving the mgnitude of the electric force (sometimes clled the Coulomb force) between two point chrges: F e k q 1 q 2 e (23.1) r 2 where k e is constnt clled the Coulomb constnt. In his experiments, Coulomb ws ble to show tht the vlue of the exponent of r ws 2 to within n uncertinty of few percent. Modern experiments hve shown tht the exponent is 2 to within n uncertinty of few prts in The vlue of the Coulomb constnt depends on the choice of units. The SI unit of chrge is the coulomb (C). The Coulomb constnt k e in SI units hs the vlue This constnt is lso written in the form k e Nm 2 /C 2 k e 1 40 where the constnt 0 (lowercse Greek epsilon) is known s the permittivity of free spce nd hs the vlue C 2 /Nm 2. The smllest unit of chrge known in nture is the chrge on n electron or proton, 1 which hs n bsolute vlue of e C Therefore, 1 C of chrge is pproximtely equl to the chrge of electrons or protons. This number is very smll when compred with the number of 1 No unit of chrge smller thn e hs been detected s free chrge; however, recent theories propose the existence of prticles clled qurks hving chrges e/3 nd 2e/3. Although there is considerble experimentl evidence for such prticles inside nucler mtter, free qurks hve never been detected. We discuss other properties of qurks in Chpter 46 of the extended version of this text.

8 23.3 Coulomb s Lw 715 TABLE 23.1 Chrge nd Mss of the Electron, Proton, nd Neutron Prticle Chrge (C) Mss (kg) Electron (e) Proton (p) Neutron (n) free electrons 2 in 1 cm 3 of copper, which is of the order of Still, 1 C is substntil mount of chrge. In typicl experiments in which rubber or glss rod is chrged by friction, net chrge of the order of 10 6 C is obtined. In other words, only very smll frction of the totl vilble chrge is trnsferred between the rod nd the rubbing mteril. The chrges nd msses of the electron, proton, nd neutron re given in Tble EXAMPLE 23.1 The Hydrogen Atom The electron nd proton of hydrogen tom re seprted (on the verge) by distnce of pproximtely m. Find the mgnitudes of the electric force nd the grvittionl force between the two prticles. Solution From Coulomb s lw, we find tht the ttrctive electric force hs the mgnitude F e k e e 2 r Nm2 ( C) 2 C 2 ( m) N Using Newton s lw of grvittion nd Tble 23.1 for the prticle msses, we find tht the grvittionl force hs the mgnitude F g G m em p r Nm2 kg 2 ( kg)( kg) ( m) N The rtio F e /F g Thus, the grvittionl force between chrged tomic prticles is negligible when compred with the electric force. Note the similrity of form of Newton s lw of grvittion nd Coulomb s lw of electric forces. Other thn mgnitude, wht is fundmentl difference between the two forces? When deling with Coulomb s lw, you must remember tht force is vector quntity nd must be treted ccordingly. Thus, the lw expressed in vector form for the electric force exerted by chrge q 1 on second chrge q 2, written F 12, is F 12 k q 1q 2 e (23.2) r 2 rˆ where rˆ is unit vector directed from q 1 to q 2, s shown in Figure Becuse the electric force obeys Newton s third lw, the electric force exerted by q 2 on q 1 is 2 A metl tom, such s copper, contins one or more outer electrons, which re wekly bound to the nucleus. When mny toms combine to form metl, the so-clled free electrons re these outer electrons, which re not bound to ny one tom. These electrons move bout the metl in mnner similr to tht of gs molecules moving in continer.

9 716 CHAPTER 23 Electric Fields r q 2 F 12 q 1 rˆ F 21 () q 1 F 21 (b) F 12 q 2 Figure 23.6 Two point chrges seprted by distnce r exert force on ech other tht is given by Coulomb s lw. The force F 21 exerted by q 2 on q 1 is equl in mgnitude nd opposite in direction to the force F 12 exerted by q 1 on q 2. () When the chrges re of the sme sign, the force is repulsive. (b) When the chrges re of opposite signs, the force is ttrctive. equl in mgnitude to the force exerted by q 1 on q 2 nd in the opposite direction; tht is, F 21 F 12. Finlly, from Eqution 23.2, we see tht if q 1 nd q 2 hve the sme sign, s in Figure 23.6, the product q 1 q 2 is positive nd the force is repulsive. If q 1 nd q 2 re of opposite sign, s shown in Figure 23.6b, the product q 1 q 2 is negtive nd the force is ttrctive. Noting the sign of the product q 1 q 2 is n esy wy of determining the direction of forces cting on the chrges. Quick Quiz 23.3 Object A hs chrge of 2 C, nd object B hs chrge of 6 C. Which sttement is true? () F AB 3F BA. (b) F AB F BA. (c) 3F AB F BA. When more thn two chrges re present, the force between ny pir of them is given by Eqution Therefore, the resultnt force on ny one of them equls the vector sum of the forces exerted by the vrious individul chrges. For exmple, if four chrges re present, then the resultnt force exerted by prticles 2, 3, nd 4 on prticle 1 is F 1 F 21 F 31 F 41 EXAMPLE 23.2 Find the Resultnt Force Consider three point chrges locted t the corners of right tringle s shown in Figure 23.7, where q 1 q C, q C, nd 0.10 m. Find the resultnt force exerted on q 3. Solution First, note the direction of the individul forces exerted by q 1 nd q 2 on q 3. The force F 23 exerted by q 2 on q 3 is ttrctive becuse q 2 nd q 3 hve opposite signs. The force F 13 exerted by q 1 on q 3 is repulsive becuse both chrges re positive. The mgnitude of F 23 is F 23 k e q 2 q Nm2 ( C)( C) C 2 (0.10 m) N Note tht becuse q 3 nd q 2 hve opposite signs, F 23 is to the left, s shown in Figure 23.7.

10 23.3 Coulomb s Lw 717 Figure 23.7 q 2 q 1 y 2 The force exerted by q 1 on q 3 is F 13. The force exerted by q 2 on q 3 is F 23. The resultnt force F 3 exerted on q 3 is the vector sum F 13 F 23. F 23 q 3 F 13 x Nm2 ( C)( C) C 2 2(0.10 m) 2 11 N The force F 13 is repulsive nd mkes n ngle of 45 with the x xis. Therefore, the x nd y components of F 13 re equl, with mgnitude given by F 13 cos N. The force F 23 is in the negtive x direction. Hence, the x nd y components of the resultnt force cting on q 3 re F 3x F 13x F N 9.0 N 1.1 N F 3y F 13y 7.9 N We cn lso express the resultnt force cting on q 3 in unitvector form s F 3 (1.1i 7.9j) N The mgnitude of the force exerted by q 1 on q 3 is F 13 k q 1 q 3 e (!2) 2 Exercise force F 3. Answer Find the mgnitude nd direction of the resultnt 8.0 N t n ngle of 98 with the x xis. EXAMPLE 23.3 Where Is the Resultnt Force Zero? Three point chrges lie long the x xis s shown in Figure The positive chrge q C is t x 2.00 m, the positive chrge q C is t the origin, nd the resultnt force cting on q 3 is zero. Wht is the x coordinte of q 3? Solution Becuse q 3 is negtive nd q 1 nd q 2 re positive, the forces F 13 nd F 23 re both ttrctive, s indicted in Figure From Coulomb s lw, F 13 nd F 23 hve mgnitudes F 13 k e q 1 q 3 (2.00 x) 2 F 23 k q 2 q 3 e x 2 For the resultnt force on q 3 to be zero, F 23 must be equl in mgnitude nd opposite in direction to F 13, or k e q 2 q 3 x 2 k e q 1 q 3 (2.00 x) 2 Noting tht k e nd q 3 re common to both sides nd so cn be dropped, we solve for x nd find tht ( x x 2 )( C) x 2 ( C) Solving this qudrtic eqution for x, we find tht x m. Figure 23.8 (2.00 x) 2 q 2 x 2 q 1 Why is the negtive root not cceptble? x 2.00 m 2.00 x q 2 F 23 q 3 F 13 q 1 Three point chrges re plced long the x xis. If the net force cting on q 3 is zero, then the force F 13 exerted by q 1 on q 3 must be equl in mgnitude nd opposite in direction to the force F 23 exerted by q 2 on q 3. x EXAMPLE 23.4 Find the Chrge on the Spheres Two identicl smll chrged spheres, ech hving mss of kg, hng in equilibrium s shown in Figure The length of ech string is 0.15 m, nd the ngle is 5.0. Find the mgnitude of the chrge on ech sphere. Solution From the right tringle shown in Figure 23.9, we see tht sin /L. Therefore, L sin (0.15 m)sin m The seprtion of the spheres is m. The forces cting on the left sphere re shown in Figure 23.9b. Becuse the sphere is in equilibrium, the forces in the

11 718 CHAPTER 23 Electric Fields horizontl nd verticl directions must seprtely dd up to zero: (1) F x T sin F e 0 (2) F y T cos mg 0 From Eqution (2), we see tht T mg /cos ; thus, T cn be q L Figure 23.9 θ () θ L = 0.15 m θ = 5.0 L q T cos θ () Two identicl spheres, ech crrying the sme chrge q, suspended in equilibrium. (b) The free-body digrm for the sphere on the left. F e (b) θ mg T θ T sin θ eliminted from Eqution (1) if we mke this substitution. This gives vlue for the mgnitude of the electric force F e : (3) F e mg tn From Coulomb s lw (Eq. 23.1), the mgnitude of the electric force is where r m nd q is the mgnitude of the chrge on ech sphere. (Note tht the term q 2 rises here becuse the chrge is the sme on both spheres.) This eqution cn be solved for q 2 to give Exercise If the chrge on the spheres were negtive, how mny electrons would hve to be dded to them to yield net chrge of C? Answer q 2 F er 2 q ( kg)(9.80 m/s 2 )tn N F e k e ( N)(0.026 m) 2 k e Nm 2 /C C electrons. q 2 r 2 QuickLb For this experiment you need two 20-cm strips of trnsprent tpe (mss of ech 65 mg). Fold bout 1 cm of tpe over t one end of ech strip to crete hndle. Press both pieces of tpe side by side onto tble top, rubbing your finger bck nd forth cross the strips. Quickly pull the strips off the surfce so tht they become chrged. Hold the tpe hndles together nd the strips will repel ech other, forming n inverted V shpe. Mesure the ngle between the pieces, nd estimte the excess chrge on ech strip. Assume tht the chrges ct s if they were locted t the center of mss of ech strip. Q Figure A smll positive test chrge q 0 plced ner n object crrying much lrger positive chrge Q experiences n electric field E directed s shown. q 0 E THE ELECTRIC FIELD Two field forces hve been introduced into our discussions so fr the grvittionl force nd the electric force. As pointed out erlier, field forces cn ct through spce, producing n effect even when no physicl contct between the objects occurs. The grvittionl field g t point in spce ws defined in Section 14.6 to be equl to the grvittionl force F g cting on test prticle of mss m divided by tht mss: g F g /m. A similr pproch to electric forces ws developed by Michel Frdy nd is of such prcticl vlue tht we shll devote much ttention to it in the next severl chpters. In this pproch, n electric field is sid to exist in the region of spce round chrged object. When nother chrged object enters this electric field, n electric force cts on it. As n exmple, consider Figure 23.10, which shows smll positive test chrge q 0 plced ner second object crrying much greter positive chrge Q. We define the strength (in other words, the mgnitude) of the electric field t the loction of the test chrge to be the electric force per unit chrge, or to be more specific

12 23.4 The Electric Field 719 the electric field E t point in spce is defined s the electric force F e cting on positive test chrge q 0 plced t tht point divided by the mgnitude of the test chrge: E F e q 0 (23.3) Definition of electric field Note tht E is the field produced by some chrge externl to the test chrge it is not the field produced by the test chrge itself. Also, note tht the existence of n electric field is property of its source. For exmple, every electron comes with its own electric field. The vector E hs the SI units of newtons per coulomb (N/C), nd, s Figure shows, its direction is the direction of the force positive test chrge experiences when plced in the field. We sy tht n electric field exists t point if test chrge t rest t tht point experiences n electric force. Once the mgnitude nd direction of the electric field re known t some point, the electric force exerted on ny chrged prticle plced t tht point cn be clculted from This drmtic photogrph cptures lightning bolt striking tree ner some rurl homes.

13 720 CHAPTER 23 Electric Fields TABLE 23.2 Source Typicl Electric Field Vlues E (N/C) Fluorescent lighting tube 10 Atmosphere (fir wether) 100 Blloon rubbed on hir Atmosphere (under thundercloud) Photocopier Sprk in ir Ner electron in hydrogen tom q 0 q 0 >>q 0 () Figure (b) () For smll enough test chrge q 0, the chrge distribution on the sphere is undisturbed. (b) When the test chrge q 0 is greter, the chrge distribution on the sphere is disturbed s the result of the proximity of q 0. q q rˆ rˆ Figure r () E (b) q 0 A test chrge q 0 t point P is distnce r from point chrge q. () If q is positive, then the electric field t P points rdilly outwrd from q. (b) If q is negtive, then the electric field t P points rdilly inwrd towrd q. P q 0 P E Eqution Furthermore, the electric field is sid to exist t some point (even empty spce) regrdless of whether test chrge is locted t tht point. (This is nlogous to the grvittionl field set up by ny object, which is sid to exist t given point regrdless of whether some other object is present t tht point to feel the field.) The electric field mgnitudes for vrious field sources re given in Tble When using Eqution 23.3, we must ssume tht the test chrge q 0 is smll enough tht it does not disturb the chrge distribution responsible for the electric field. If vnishingly smll test chrge q 0 is plced ner uniformly chrged metllic sphere, s shown in Figure 23.11, the chrge on the metllic sphere, which produces the electric field, remins uniformly distributed. If the test chrge is gret enough (q 0 W q 0 ), s shown in Figure 23.11b, the chrge on the metllic sphere is redistributed nd the rtio of the force to the test chrge is different: (F e /q 0 F e /q 0 ). Tht is, becuse of this redistribution of chrge on the metllic sphere, the electric field it sets up is different from the field it sets up in the presence of the much smller q 0. To determine the direction of n electric field, consider point chrge q locted distnce r from test chrge q 0 locted t point P, s shown in Figure According to Coulomb s lw, the force exerted by q on the test chrge is F e k e qq 0 r 2 rˆ where rˆ is unit vector directed from q towrd q 0. Becuse the electric field t P, the position of the test chrge, is defined by E F e /q 0, we find tht t P, the electric field creted by q is E k q e (23.4) r 2 rˆ If q is positive, s it is in Figure 23.12, the electric field is directed rdilly outwrd from it. If q is negtive, s it is in Figure 23.12b, the field is directed towrd it. To clculte the electric field t point P due to group of point chrges, we first clculte the electric field vectors t P individully using Eqution 23.4 nd then dd them vectorilly. In other words, t ny point P, the totl electric field due to group of chrges equls the vector sum of the electric fields of the individul chrges. This superposition principle pplied to fields follows directly from the superposition property of electric forces. Thus, the electric field of group of chrges cn

14 23.4 The Electric Field 721 This metllic sphere is chrged by genertor so tht it crries net electric chrge. The high concentrtion of chrge on the sphere cretes strong electric field round the sphere. The chrges then lek through the gs surrounding the sphere, producing pink glow. be expressed s q E k e i (23.5) i r 2 rˆi i where r i is the distnce from the ith chrge q i to the point P (the loction of the test chrge) nd rˆi is unit vector directed from q i towrd P. Quick Quiz 23.4 A chrge of 3 C is t point P where the electric field is directed to the right nd hs mgnitude of N/C. If the chrge is replced with 3-C chrge, wht hppens to the electric field t P? EXAMPLE 23.5 Electric Field Due to Two Chrges A chrge q C is locted t the origin, nd second chrge q C is locted on the x xis, 0.30 m from the origin (Fig ). Find the electric field t the point P, which hs coordintes (0, 0.40) m. Solution First, let us find the mgnitude of the electric field t P due to ech chrge. The fields E 1 due to the 7.0-C chrge nd E 2 due to the 5.0-C chrge re shown in Figure Their mgnitudes re E 1 k q 1 e r Nm2 ( C) C 2 (0.40 m) m N/C E 2 k q 2 e r Nm2 ( C) C 2 (0.50 m) N/C The vector E 1 hs only y component. The vector E 2 hs n x component given by E 2 cos 3 nd negtive y component given by E 2 sin 4 5 E 2. 5 E 2 Hence, we cn express the vectors s Figure E 1 P y q 1 φ θ E E m θ 0.30 m The totl electric field E t P equls the vector sum E 1 E 2, where E 1 is the field due to the positive chrge q 1 nd E 2 is the field due to the negtive chrge q 2. q 2 x

15 722 CHAPTER 23 Electric Fields E j N/C E 2 ( i j) N/C The resultnt field E t P is the superposition of E 1 nd E 2 : E E ( i E 2 j) N/C From this result, we find tht E hs mgnitude of N/C nd mkes n ngle of 66 with the positive x xis. Exercise Find the electric force exerted on chrge of C locted t P. Answer N in the sme direction s E. EXAMPLE 23.6 Electric Field of Dipole An electric dipole is defined s positive chrge q nd negtive chrge q seprted by some distnce. For the dipole shown in Figure 23.14, find the electric field E t P due to the chrges, where P is distnce y W from the origin. Solution At P, the fields E 1 nd E 2 due to the two chrges re equl in mgnitude becuse P is equidistnt from the chrges. The totl field is E E 1 E 2, where The y components of E 1 nd E 2 cncel ech other, nd the x components dd becuse they re both in the positive x direction. Therefore, E is prllel to the x xis nd hs mgnitude equl to 2E 1 cos. From Figure we see tht cos Therefore, /r /(y 2 2 ) 1/2. E 2E 1 cos 2k e k e E 1 E 2 k e q r 2 2q (y 2 2 ) 3/2 k e q (y 2 2 ) Becuse y W, we cn neglect 2 nd write E k e 2q y 3 q y 2 2 (y 2 2 ) 1/2 Thus, we see tht, t distnces fr from dipole but long the perpendiculr bisector of the line joining the two chrges, the mgnitude of the electric field creted by the dipole vries s 1/r 3, wheres the more slowly vrying field of point chrge vries s 1/r 2 (see Eq. 23.4). This is becuse t distnt points, the fields of the two chrges of equl mgnitude nd opposite sign lmost cncel ech other. The 1/r 3 vrition in E for the dipole lso is obtined for distnt point long the x xis (see Problem 21) nd for ny generl distnt point. The electric dipole is good model of mny molecules, such s hydrochloric cid (HCl). As we shll see in lter chpters, neutrl toms nd molecules behve s dipoles when plced in n externl electric field. Furthermore, mny molecules, such s HCl, re permnent dipoles. The effect of such dipoles on the behvior of mterils subjected to electric fields is discussed in Chpter 26. Figure q r θ P y y θ θ θ E 1 E 2 E q x The totl electric field E t P due to two chrges of equl mgnitude nd opposite sign (n electric dipole) equls the vector sum E 1 E 2. The field E 1 is due to the positive chrge q, nd E 2 is the field due to the negtive chrge q ELECTRIC FIELD OF A CONTINUOUS CHARGE DISTRIBUTION Very often the distnces between chrges in group of chrges re much smller thn the distnce from the group to some point of interest (for exmple, point where the electric field is to be clculted). In such situtions, the system of

16 23.5 Electric Field of Continuous Chrge Distribution 723 chrges is smered out, or continuous. Tht is, the system of closely spced chrges is equivlent to totl chrge tht is continuously distributed long some line, over some surfce, or throughout some volume. To evlute the electric field creted by continuous chrge distribution, we use the following procedure: First, we divide the chrge distribution into smll elements, ech of which contins smll chrge q, s shown in Figure Next, we use Eqution 23.4 to clculte the electric field due to one of these elements t point P. Finlly, we evlute the totl field t P due to the chrge distribution by summing the contributions of ll the chrge elements (tht is, by pplying the superposition principle). The electric field t P due to one element crrying chrge q is where r is the distnce from the element to point P nd rˆ is unit vector directed from the chrge element towrd P. The totl electric field t P due to ll elements in the chrge distribution is pproximtely where the index i refers to the ith element in the distribution. Becuse the chrge distribution is pproximtely continuous, the totl field t P in the limit q i : 0 is E k e lim q i :0 i E k e q r 2 rˆ E k e i q i r 2 rˆi i q i r 2 rˆi k e i dq r 2 rˆ (23.6) where the integrtion is over the entire chrge distribution. This is vector opertion nd must be treted ppropritely. We illustrte this type of clcultion with severl exmples, in which we ssume the chrge is uniformly distributed on line, on surfce, or throughout volume. When performing such clcultions, it is convenient to use the concept of chrge density long with the following nottions: A continuous chrge distribution E r P Figure q The electric field t P due to continuous chrge distribution is the vector sum of the fields E due to ll the elements q of the chrge distribution. Electric field of continuous chrge distribution rˆ If chrge Q is uniformly distributed throughout volume V, the volume chrge density is defined by Q V Volume chrge density where hs units of coulombs per cubic meter (C/m 3 ). If chrge Q is uniformly distributed on surfce of re A, the surfce chrge density (lowercse Greek sigm) is defined by Q A Surfce chrge density where hs units of coulombs per squre meter (C/m 2 ). If chrge Q is uniformly distributed long line of length, the liner chrge density is defined by Q Liner chrge density where hs units of coulombs per meter (C/m).

17 724 CHAPTER 23 Electric Fields If the chrge is nonuniformly distributed over volume, surfce, or line, we hve to express the chrge densities s dq dv dq where dq is the mount of chrge in smll volume, surfce, or length element. da dq d EXAMPLE 23.7 The Electric Field Due to Chrged Rod A rod of length hs uniform positive chrge per unit length nd totl chrge Q. Clculte the electric field t point P tht is locted long the long xis of the rod nd distnce from one end (Fig ). Solution Let us ssume tht the rod is lying long the x xis, tht dx is the length of one smll segment, nd tht dq is the chrge on tht segment. Becuse the rod hs chrge per unit length, the chrge dq on the smll segment is dq The field d E due to this segment t P is in the negtive x direction (becuse the source of the field crries positive chrge Q ), nd its mgnitude is dx. de k dq e x 2 k e dx x 2 Becuse every other element lso produces field in the negtive x direction, the problem of summing their contributions is prticulrly simple in this cse. The totl field t P due to ll segments of the rod, which re t different distnces from P, is given by Eqution 23.6, which in this cse becomes 3 E k e dx where the limits on the integrl extend from one end of the rod (x ) to the other (x ). The constnts k e nd cn be removed from the integrl to yield x 2 E k e dx x 2 k e 1 x k e 1 1 k e Q ( ) where we hve used the fct tht the totl chrge Q. If P is fr from the rod ( W ), then the in the denomintor cn be neglected, nd E k e Q / 2. This is just the form you would expect for point chrge. Therefore, t lrge vlues of /, the chrge distribution ppers to be point chrge of mgnitude Q. The use of the limiting technique (/ : ) often is good method for checking theoreticl formul. de P Figure y x dx dq = λdxλ The electric field t P due to uniformly chrged rod lying long the x xis. The mgnitude of the field t P due to the segment of chrge dq is k e dq/x 2. The totl field t P is the vector sum over ll segments of the rod. x EXAMPLE 23.8 The Electric Field of Uniform Ring of Chrge A ring of rdius crries uniformly distributed positive totl chrge Q. Clculte the electric field due to the ring t point P lying distnce x from its center long the centrl xis perpendiculr to the plne of the ring (Fig ). Solution The mgnitude of the electric field t P due to the segment of chrge dq is de k e dq r 2 This field hs n x component de x de cos long the xis nd component de perpendiculr to the xis. As we see in Figure 23.17b, however, the resultnt field t P must lie long the x xis becuse the perpendiculr components of ll the 3 It is importnt tht you understnd how to crry out integrtions such s this. First, express the chrge element dq in terms of the other vribles in the integrl (in this exmple, there is one vrible, x, nd so we mde the chnge dq The integrl must be over sclr quntities; therefore, you must express the electric field in terms of components, if necessry. (In this exmple the field hs only n x component, so we do not bother with this detil.) Then, reduce your expression to n integrl over single vrible (or to multiple integrls, ech over single vrible). In exmples tht hve sphericl or cylindricl symmetry, the single vrible will be rdil coordinte. dx).

18 If we consider the disk s set of concentric rings, we cn use our result from Exmple 23.8 which gives the field creted by ring of rdius nd sum the contrivrious chrge segments sum to zero. Tht is, the perpendiculr component of the field creted by ny chrge element is cnceled by the perpendiculr component creted by n element on the opposite side of the ring. Becuse r (x 2 2 ) 1/2 nd cos x/r, we find tht de x de cos k e dq r 2 x r All segments of the ring mke the sme contribution to the field t P becuse they re ll equidistnt from this point. Thus, we cn integrte to obtin the totl field t P : Figure dq 23.5 Electric Field of Continuous Chrge Distribution 725 k e x (x 2 2 ) 3/2 dq x () r θ P de de x de This result shows tht the field is zero t x 0. Does this finding surprise you? Exercise E x k e x (x 2 2 ) 3/2 dq k e x (x 2 2 ) 3/2 dq k e x (x 2 2 ) 3/2 Q Show tht t gret distnces from the ring (x W ) the electric field long the xis shown in Figure pproches tht of point chrge of mgnitude Q. A uniformly chrged ring of rdius. () The field t P on the x xis due to n element of chrge dq. (b) The totl electric field t P is long the x xis. The perpendiculr component of the field t P due to segment 1 is cnceled by the perpendiculr component due to segment (b) θ de 1 de 2 EXAMPLE 23.9 The Electric Field of Uniformly Chrged Disk A disk of rdius R hs uniform surfce chrge density. Clculte the electric field t point P tht lies long the centrl perpendiculr xis of the disk nd distnce x from the center of the disk (Fig ). Solution Figure r R dr A uniformly chrged disk of rdius R. The electric field t n xil point P is directed long the centrl xis, perpendiculr to the plne of the disk. dq x P butions of ll rings mking up the disk. By symmetry, the field t n xil point must be long the centrl xis. The ring of rdius r nd width dr shown in Figure hs surfce re equl to 2r dr. The chrge dq on this ring is equl to the re of the ring multiplied by the surfce chrge density: dq 2r dr. Using this result in the eqution given for E x in Exmple 23.8 (with replced by r), we hve for the field due to the ring de To obtin the totl field t P, we integrte this expression over the limits r 0 to r R, noting tht x is constnt. This gives E k e x R k e x (x 2 r 2 ) 3/2 (2r dr) 0 k e x R (x 2 r 2 ) 3/2 d(r 2 ) 0 k e x (x 2 r 2 ) 1/2 1/2 R 0 2k e x x 2r dr (x 2 r 2 ) 3/2 x (x 2 R 2 ) 1/2

19 726 CHAPTER 23 Electric Fields This result is vlid for ll vlues of x. We cn clculte the field close to the disk long the xis by ssuming tht R W x ; thus, the expression in prentheses reduces to unity: 0 1/(4k e ) where is the permittivity of free spce. As we shll find in the next chpter, we obtin the sme result for the field creted by uniformly chrged infinite sheet. E 2k e 2 0 A Figure Electric field lines penetrting two surfces. The mgnitude of the field is greter on surfce A thn on surfce B. B ELECTRIC FIELD LINES A convenient wy of visulizing electric field ptterns is to drw lines tht follow the sme direction s the electric field vector t ny point. These lines, clled electric field lines, re relted to the electric field in ny region of spce in the following mnner: The electric field vector E is tngent to the electric field line t ech point. The number of lines per unit re through surfce perpendiculr to the lines is proportionl to the mgnitude of the electric field in tht region. Thus, E is gret when the field lines re close together nd smll when they re fr prt. These properties re illustrted in Figure The density of lines through surfce A is greter thn the density of lines through surfce B. Therefore, the electric field is more intense on surfce A thn on surfce B. Furthermore, the fct tht the lines t different loctions point in different directions indictes tht the field is nonuniform. Representtive electric field lines for the field due to single positive point chrge re shown in Figure Note tht in this two-dimensionl drwing we show only the field lines tht lie in the plne contining the point chrge. The lines re ctully directed rdilly outwrd from the chrge in ll directions; thus, insted of the flt wheel of lines shown, you should picture n entire sphere of lines. Becuse positive test chrge plced in this field would be repelled by the positive point chrge, the lines re directed rdilly wy from the positive point q q () Figure (b) The electric field lines for point chrge. () For positive point chrge, the lines re directed rdilly outwrd. (b) For negtive point chrge, the lines re directed rdilly inwrd. Note tht the figures show only those field lines tht lie in the plne contining the chrge. (c) The drk res re smll pieces of thred suspended in oil, which lign with the electric field produced by smll chrged conductor t the center. (c)

20 23.6 Electric Field Lines 727 chrge. The electric field lines representing the field due to single negtive point chrge re directed towrd the chrge (Fig b). In either cse, the lines re long the rdil direction nd extend ll the wy to infinity. Note tht the lines become closer together s they pproch the chrge; this indictes tht the strength of the field increses s we move towrd the source chrge. The rules for drwing electric field lines re s follows: The lines must begin on positive chrge nd terminte on negtive chrge. The number of lines drwn leving positive chrge or pproching negtive chrge is proportionl to the mgnitude of the chrge. No two field lines cn cross. Rules for drwing electric field lines Is this visuliztion of the electric field in terms of field lines consistent with Eqution 23.4, the expression we obtined for E using Coulomb s lw? To nswer this question, consider n imginry sphericl surfce of rdius r concentric with point chrge. From symmetry, we see tht the mgnitude of the electric field is the sme everywhere on the surfce of the sphere. The number of lines N tht emerge from the chrge is equl to the number tht penetrte the sphericl surfce. Hence, the number of lines per unit re on the sphere is N/4r 2 (where the surfce re of the sphere is 4r 2 ). Becuse E is proportionl to the number of lines per unit re, we see tht E vries s 1/r 2 ; this finding is consistent with Eqution As we hve seen, we use electric field lines to qulittively describe the electric field. One problem with this model is tht we lwys drw finite number of lines from (or to) ech chrge. Thus, it ppers s if the field cts only in certin directions; this is not true. Insted, the field is continuous tht is, it exists t every point. Another problem ssocited with this model is the dnger of gining the wrong impression from two-dimensionl drwing of field lines being used to describe three-dimensionl sitution. Be wre of these shortcomings every time you either drw or look t digrm showing electric field lines. We choose the number of field lines strting from ny positively chrged object to be Cq nd the number of lines ending on ny negtively chrged object to be C q, where C is n rbitrry proportionlity constnt. Once C is chosen, the number of lines is fixed. For exmple, if object 1 hs chrge Q 1 nd object 2 hs chrge Q 2, then the rtio of number of lines is N 2 /N 1 Q 2 /Q 1. The electric field lines for two point chrges of equl mgnitude but opposite signs (n electric dipole) re shown in Figure Becuse the chrges re of equl mgnitude, the number of lines tht begin t the positive chrge must equl the number tht terminte t the negtive chrge. At points very ner the chrges, the lines re nerly rdil. The high density of lines between the chrges indictes region of strong electric field. Figure shows the electric field lines in the vicinity of two equl positive point chrges. Agin, the lines re nerly rdil t points close to either chrge, nd the sme number of lines emerge from ech chrge becuse the chrges re equl in mgnitude. At gret distnces from the chrges, the field is pproximtely equl to tht of single point chrge of mgnitude 2q. Finlly, in Figure we sketch the electric field lines ssocited with positive chrge 2q nd negtive chrge q. In this cse, the number of lines leving 2q is twice the number terminting t q. Hence, only hlf of the lines tht leve the positive chrge rech the negtive chrge. The remining hlf terminte Figure () (b) () The electric field lines for two point chrges of equl mgnitude nd opposite sign (n electric dipole). The number of lines leving the positive chrge equls the number terminting t the negtive chrge. (b) The drk lines re smll pieces of thred suspended in oil, which lign with the electric field of dipole.

21 728 CHAPTER 23 Electric Fields B A C () (b) Figure () The electric field lines for two positive point chrges. (The loctions A, B, nd C re discussed in Quick Quiz 23.5.) (b) Pieces of thred suspended in oil, which lign with the electric field creted by two equl-mgnitude positive chrges. 2q q on negtive chrge we ssume to be t infinity. At distnces tht re much greter thn the chrge seprtion, the electric field lines re equivlent to those of single chrge q. Figure The electric field lines for point chrge 2q nd second point chrge q. Note tht two lines leve 2q for every one tht termintes on q. Quick Quiz 23.5 Rnk the mgnitude of the electric field t points A, B, nd C shown in Figure (gretest mgnitude first) MOTION OF CHARGED PARTICLES IN A UNIFORM ELECTRIC FIELD When prticle of chrge q nd mss m is plced in n electric field E, the electric force exerted on the chrge is qe. If this is the only force exerted on the prticle, it must be the net force nd so must cuse the prticle to ccelerte. In this cse, Newton s second lw pplied to the prticle gives F e qe m The ccelertion of the prticle is therefore qe m (23.7) If E is uniform (tht is, constnt in mgnitude nd direction), then the ccelertion is constnt. If the prticle hs positive chrge, then its ccelertion is in the direction of the electric field. If the prticle hs negtive chrge, then its ccelertion is in the direction opposite the electric field. EXAMPLE An Accelerting Positive Chrge A positive point chrge q of mss m is relesed from rest in uniform electric field E directed long the x xis, s shown in Figure Describe its motion. Solution The ccelertion is constnt nd is given by qe/m. The motion is simple liner motion long the x xis. Therefore, we cn pply the equtions of kinemtics in one

22 23.7 Motion of Chrged Prticles in Uniform Electric Field 729 dimension (see Chpter 2): x f x i v xi t 1 2 xt 2 v xf v xi x t v xf 2 v xi 2 2 x (x f x i ) Tking x i 0 nd v xi 0, we hve x f 1 2 xt 2 qe 2m t 2 v xf x t qe m t v 2 xf 2 x x f 2qE m x f The kinetic energy of the chrge fter it hs moved distnce x x f x i is K 1 2 mv m 2qE m x qex We cn lso obtin this result from the workkinetic energy theorem becuse the work done by the electric force is F e x qex nd W K. Figure E v v = 0 q A positive point chrge q in uniform electric field E undergoes constnt ccelertion in the direction of the field. x The electric field in the region between two oppositely chrged flt metllic pltes is pproximtely uniform (Fig ). Suppose n electron of chrge e is projected horizontlly into this field with n initil velocity v i i. Becuse the electric field E in Figure is in the positive y direction, the ccelertion of the electron is in the negtive y direction. Tht is, ee m j (23.8) Becuse the ccelertion is constnt, we cn pply the equtions of kinemtics in two dimensions (see Chpter 4) with v xi v i nd v yi 0. After the electron hs been in the electric field for time t, the components of its velocity re v x v i constnt (23.9) v y y t ee m t (23.10) v i i (0, 0) (x,y) E y v x Figure An electron is projected horizontlly into uniform electric field produced by two chrged pltes. The electron undergoes downwrd ccelertion (opposite E), nd its motion is prbolic while it is between the pltes.

23 730 CHAPTER 23 Electric Fields Its coordintes fter time t in the field re x v i t y 1 2 yt ee m t 2 (23.11) (23.12) Substituting the vlue t x/v i from Eqution into Eqution 23.12, we see tht y is proportionl to x 2. Hence, the trjectory is prbol. After the electron leves the field, it continues to move in stright line in the direction of v in Figure 23.25, obeying Newton s first lw, with speed v v i. Note tht we hve neglected the grvittionl force cting on the electron. This is good pproximtion when we re deling with tomic prticles. For n electric field of 10 4 N/C, the rtio of the mgnitude of the electric force ee to the mgnitude of the grvittionl force mg is of the order of for n electron nd of the order of for proton. EXAMPLE An Accelerted Electron An electron enters the region of uniform electric field s shown in Figure 23.25, with v i m/s nd E 200 N/C. The horizontl length of the pltes is m. () Find the ccelertion of the electron while it is in the electric field. Solution The chrge on the electron hs n bsolute vlue of C, nd m kg. Therefore, Eqution 23.8 gives ee m j ( C)(200 N/C) kg j m/s 2 (b) Find the time it tkes the electron to trvel through the field. j (c) Wht is the verticl displcement y of the electron while it is in the field? Solution Using Eqution nd the results from prts () nd (b), we find tht If the seprtion between the pltes is less thn this, the electron will strike the positive plte. Exercise the field. t v i y 1 2 yt ( m/s 2 )( s) m m m/s 1.95 cm s Find the speed of the electron s it emerges from Solution The horizontl distnce cross the field is m. Using Eqution with x, we find tht the time spent in the electric field is Answer m/s. The Cthode Ry Tube The exmple we just worked describes portion of cthode ry tube (CRT). This tube, illustrted in Figure 23.26, is commonly used to obtin visul disply of electronic informtion in oscilloscopes, rdr systems, television receivers, nd computer monitors. The CRT is vcuum tube in which bem of electrons is ccelerted nd deflected under the influence of electric or mgnetic fields. The electron bem is produced by n ssembly clled n electron gun locted in the neck of the tube. These electrons, if left undisturbed, trvel in stright-line pth until they strike the front of the CRT, the screen, which is coted with mteril tht emits visible light when bombrded with electrons. In n oscilloscope, the electrons re deflected in vrious directions by two sets of pltes plced t right ngles to ech other in the neck of the tube. (A television

24 Summry 731 Electron gun C A Verticl deflection pltes Verticl input Horizontl deflection pltes Electron bem Horizontl input Fluorescent screen Figure Schemtic digrm of cthode ry tube. Electrons leving the hot cthode C re ccelerted to the node A. In ddition to ccelerting electrons, the electron gun is lso used to focus the bem of electrons, nd the pltes deflect the bem. CRT steers the bem with mgnetic field, s discussed in Chpter 29.) An externl electric circuit is used to control the mount of chrge present on the pltes. The plcing of positive chrge on one horizontl plte nd negtive chrge on the other cretes n electric field between the pltes nd llows the bem to be steered from side to side. The verticl deflection pltes ct in the sme wy, except tht chnging the chrge on them deflects the bem verticlly. SUMMARY Electric chrges hve the following importnt properties: Unlike chrges ttrct one nother, nd like chrges repel one nother. Chrge is conserved. Chrge is quntized tht is, it exists in discrete pckets tht re some integrl multiple of the electronic chrge. Conductors re mterils in which chrges move freely. Insultors re mterils in which chrges do not move freely. Coulomb s lw sttes tht the electric force exerted by chrge q 1 on second chrge q 2 is F 12 k q 1q 2 e (23.2) r 2 rˆ where r is the distnce between the two chrges nd rˆ is unit vector directed from q 1 to q 2. The constnt k e, clled the Coulomb constnt, hs the vlue k e Nm 2 /C 2. The smllest unit of chrge known to exist in nture is the chrge on n electron or proton, e C. The electric field E t some point in spce is defined s the electric force F e tht cts on smll positive test chrge plced t tht point divided by the mgnitude of the test chrge q 0 : E F e (23.3) q 0 At distnce r from point chrge q, the electric field due to the chrge is given by E k q e (23.4) r 2 rˆ where rˆ is unit vector directed from the chrge to the point in question. The

25 732 CHAPTER 23 Electric Fields electric field is directed rdilly outwrd from positive chrge nd rdilly inwrd towrd negtive chrge. The electric field due to group of point chrges cn be obtined by using the superposition principle. Tht is, the totl electric field t some point equls the vector sum of the electric fields of ll the chrges: E k e (23.5) i r 2 rˆi i The electric field t some point of continuous chrge distribution is E k e dq (23.6) r 2 rˆ where dq is the chrge on one element of the chrge distribution nd r is the distnce from the element to the point in question. Electric field lines describe n electric field in ny region of spce. The number of lines per unit re through surfce perpendiculr to the lines is proportionl to the mgnitude of E in tht region. A chrged prticle of mss m nd chrge q moving in n electric field E hs n ccelertion q i qe m (23.7) Problem-Solving Hints Finding the Electric Field Units: In clcultions using the Coulomb constnt k e (1/40), chrges must be expressed in coulombs nd distnces in meters. Clculting the electric field of point chrges: To find the totl electric field t given point, first clculte the electric field t the point due to ech individul chrge. The resultnt field t the point is the vector sum of the fields due to the individul chrges. Continuous chrge distributions: When you re confronted with problems tht involve continuous distribution of chrge, the vector sums for evluting the totl electric field t some point must be replced by vector integrls. Divide the chrge distribution into infinitesiml pieces, nd clculte the vector sum by integrting over the entire chrge distribution. You should review Exmples 23.7 through Symmetry: With both distributions of point chrges nd continuous chrge distributions, tke dvntge of ny symmetry in the system to simplify your clcultions. QUESTIONS 1. Sprks re often observed (or herd) on dry dy when clothes re removed in the drk. Explin. 2. Explin from n tomic viewpoint why chrge is usully trnsferred by electrons. 3. A blloon is negtively chrged by rubbing nd then clings to wll. Does this men tht the wll is positively chrged? Why does the blloon eventully fll? 4. A light, unchrged metllic sphere suspended from thred is ttrcted to chrged rubber rod. After touching the rod, the sphere is repelled by the rod. Explin.

26 Problems Explin wht is ment by the term neutrl tom. 6. Why do some clothes cling together nd to your body fter they re removed from dryer? 7. A lrge metllic sphere insulted from ground is chrged with n electrosttic genertor while person stnding on n insulting stool holds the sphere. Why is it sfe to do this? Why wouldn t it be sfe for nother person to touch the sphere fter it hs been chrged? 8. Wht re the similrities nd differences between Newton s lw of grvittion, F g Gm 1 m 2 /r 2, nd Coulomb s lw, F e k e q 1 q 2 /r 2? 9. Assume tht someone proposes theory tht sttes tht people re bound to the Erth by electric forces rther thn by grvity. How could you prove this theory wrong? 10. How would you experimentlly distinguish n electric field from grvittionl field? 11. Would life be different if the electron were positively chrged nd the proton were negtively chrged? Does the choice of signs hve ny bering on physicl nd chemicl interctions? Explin. 12. When defining the electric field, why is it necessry to specify tht the mgnitude of the test chrge be very smll (tht is, why is it necessry to tke the limit of F e /q s q : 0)? 13. Two chrged conducting spheres, ech of rdius, re seprted by distnce r 2. Is the force on either sphere given by Coulomb s lw? Explin. (Hint: Refer to Chpter 14 on grvittion.) 14. When is it vlid to pproximte chrge distribution by point chrge? 15. Is it possible for n electric field to exist in empty spce? Explin. 16. Explin why electric field lines never cross. (Hint: E must hve unique direction t ll points.) 17. A free electron nd free proton re plced in n identicl electric field. Compre the electric forces on ech prticle. Compre their ccelertions. 18. Explin wht hppens to the mgnitude of the electric field of point chrge s r pproches zero. 19. A negtive chrge is plced in region of spce where the electric field is directed verticlly upwrd. Wht is the direction of the electric force experienced by this chrge? 20. A chrge 4q is distnce r from chrge q. Compre the number of electric field lines leving the chrge 4q with the number entering the chrge q. 21. In Figure 23.23, where do the extr lines leving the chrge 2q end? 22. Consider two equl point chrges seprted by some distnce d. At wht point (other thn ) would third test chrge experience no net force? 23. A negtive point chrge q is plced t the point P ner the positively chrged ring shown in Figure If x V, describe the motion of the point chrge if it is relesed from rest. 24. Explin the differences between liner, surfce, nd volume chrge densities, nd give exmples of when ech would be used. 25. If the electron in Figure is projected into the electric field with n rbitrry velocity v i (t n ngle to E), will its trjectory still be prbolic? Explin. 26. It hs been reported tht in some instnces people ner where lightning bolt strikes the Erth hve hd their clothes thrown off. Explin why this might hppen. 27. Why should ground wire be connected to the metllic support rod for television ntenn? 28. A light strip of luminum foil is drped over wooden rod. When rod crrying positive chrge is brought close to the foil, the two prts of the foil stnd prt. Why? Wht kind of chrge is on the foil? 29. Why is it more difficult to chrge n object by rubbing on humid dy thn on dry dy? PROBLEMS 1, 2, 3 = strightforwrd, intermedite, chllenging = full solution vilble in the Student Solutions Mnul nd Study Guide WEB = solution posted t = Computer useful in solving problem = Interctive Physics = pired numericl/symbolic problems Section 23.1 Properties of Electric Chrges Section 23.2 Insultors nd Conductors Section 23.3 Coulomb s Lw 1. () Clculte the number of electrons in smll, electriclly neutrl silver pin tht hs mss of 10.0 g. Silver hs 47 electrons per tom, nd its molr mss is g/mol. (b) Electrons re dded to the pin until the net negtive chrge is 1.00 mc. How mny electrons re dded for every 10 9 electrons lredy present? 2. () Two protons in molecule re seprted by distnce of m. Find the electric force exerted by one proton on the other. (b) How does the mgnitude of this WEB force compre with the mgnitude of the grvittionl force between the two protons? (c) Wht must be the chrge-to-mss rtio of prticle if the mgnitude of the grvittionl force between two of these prticles equls the mgnitude of the electric force between them? 3. Richrd Feynmn once sid tht if two persons stood t rm s length from ech other nd ech person hd 1% more electrons thn protons, the force of repulsion between them would be enough to lift weight equl to tht of the entire Erth. Crry out n order-ofmgnitude clcultion to substntite this ssertion. 4. Two smll silver spheres, ech with mss of 10.0 g, re seprted by 1.00 m. Clculte the frction of the elec-

27 734 CHAPTER 23 Electric Fields trons in one sphere tht must be trnsferred to the other to produce n ttrctive force of N (bout 1 ton) between the spheres. (The number of electrons per tom of silver is 47, nd the number of toms per grm is Avogdro s number divided by the molr mss of silver, g/mol.) 5. Suppose tht 1.00 g of hydrogen is seprted into electrons nd protons. Suppose lso tht the protons re plced t the Erth s north pole nd the electrons re plced t the south pole. Wht is the resulting compressionl force on the Erth? 6. Two identicl conducting smll spheres re plced with their centers m prt. One is given chrge of 12.0 nc, nd the other is given chrge of 18.0 nc. () Find the electric force exerted on one sphere by the other. (b) The spheres re connected by conducting wire. Find the electric force between the two fter equilibrium hs occurred. 7. Three point chrges re locted t the corners of n equilterl tringle, s shown in Figure P23.7. Clculte the net electric force on the 7.00-C chrge. 10. Review Problem. Two identicl point chrges ech hving chrge q re fixed in spce nd seprted by distnce d. A third point chrge Q of mss m is free to move nd lies initilly t rest on perpendiculr bisector of the two fixed chrges distnce x from the midpoint of the two fixed chrges (Fig. P23.10). () Show tht if x is smll compred with d, the motion of Q is simple hrmonic long the perpendiculr bisector. Determine the period of tht motion. (b) How fst will the chrge Q be moving when it is t the midpoint between the two fixed chrges, if initilly it is relesed t distnce x V d from the midpoint? d/2 y q Q x y 7.00 µc µ d/2 x q m 2.00 µc µ 4.00 µc µ Figure P23.7 Problems 7 nd Two smll beds hving positive chrges 3q nd q re fixed t the opposite ends of horizontl insulting rod extending from the origin to the point x d. As shown in Figure P23.8, third smll chrged bed is free to slide on the rod. At wht position is the third bed in equilibrium? Cn it be in stble equilibrium? 3q q Figure P Review Problem. In the Bohr theory of the hydrogen tom, n electron moves in circulr orbit bout proton, where the rdius of the orbit is m. () Find the electric force between the two. (b) If this force cuses the centripetl ccelertion of the electron, wht is the speed of the electron? d x Section 23.4 Figure P23.10 The Electric Field 11. Wht re the mgnitude nd direction of the electric field tht will blnce the weight of () n electron nd (b) proton? (Use the dt in Tble 23.1.) 12. An object hving net chrge of 24.0 C is plced in uniform electric field of 610 N/C tht is directed verticlly. Wht is the mss of this object if it flots in the field? 13. In Figure P23.13, determine the point (other thn infinity) t which the electric field is zero m 2.50 µc µ 6.00 µc µ Figure P An irplne is flying through thundercloud t height of m. (This is very dngerous thing to do becuse of updrfts, turbulence, nd the possibility of electric dischrge.) If there re chrge concentrtions of 40.0 C t height of m within the cloud nd of 40.0 C t height of m, wht is the electric field E t the ircrft?

28 Problems Three chrges re t the corners of n equilterl tringle, s shown in Figure P23.7. () Clculte the electric field t the position of the 2.00-C chrge due to the 7.00-C nd 4.00-C chrges. (b) Use your nswer to prt () to determine the force on the 2.00-C chrge. 16. Three point chrges re rrnged s shown in Figure P () Find the vector electric field tht the 6.00-nC nd 3.00-nC chrges together crete t the origin. (b) Find the vector force on the 5.00-nC chrge. 2q 3q q 4q y Figure P nc 3.00 nc m m Figure P nc x nents of the electric field t point (x, y) due to this chrge q re E x E y k e q(x x 0 ) [(x x 0 ) 2 (y y 0 ) 2 ] 3/2 k e q(y y 0 ) [(x x 0 ) 2 (y y 0 ) 2 ] 3/2 21. Consider the electric dipole shown in Figure P Show tht the electric field t distnt point long the x xis is E x 4k e q/x 3. WEB 17. Three equl positive chrges q re t the corners of n equilterl tringle of side, s shown in Figure P () Assume tht the three chrges together crete n electric field. Find the loction of point (other thn ) where the electric field is zero. (Hint: Sketch the field lines in the plne of the chrges.) (b) Wht re the mgnitude nd direction of the electric field t P due to the two chrges t the bse? P q q Figure P Two 2.00-C point chrges re locted on the x xis. One is t x 1.00 m, nd the other is t x 1.00 m. () Determine the electric field on the y xis t y m. (b) Clculte the electric force on 3.00-C chrge plced on the y xis t y m. 19. Four point chrges re t the corners of squre of side, s shown in Figure P () Determine the mgnitude nd direction of the electric field t the loction of chrge q. (b) Wht is the resultnt force on q? 20. A point prticle hving chrge q is locted t point (x 0, y 0 ) in the xy plne. Show tht the x nd y compo- q y q q 2 Figure P Consider n equl positive point chrges ech of mgnitude Q /n plced symmetriclly round circle of rdius R. () Clculte the mgnitude of the electric field E t point distnce x on the line pssing through the center of the circle nd perpendiculr to the plne of the circle. (b) Explin why this result is identicl to the one obtined in Exmple Consider n infinite number of identicl chrges (ech of chrge q) plced long the x xis t distnces, 2, 3, 4,...from the origin. Wht is the electric field t the origin due to this distribution? Hint: Use the fct tht Section 23.5 Electric Field of Continuous Chrge Distribution 24. A rod 14.0 cm long is uniformly chrged nd hs totl chrge of 22.0 C. Determine the mgnitude nd direction of the electric field long the xis of the rod t point 36.0 cm from its center. x 2 6

29 736 CHAPTER 23 Electric Fields WEB 25. A continuous line of chrge lies long the x xis, extending from x x 0 to positive infinity. The line crries uniform liner chrge density 0. Wht re the mgnitude nd direction of the electric field t the origin? 26. A line of chrge strts t x x 0 nd extends to positive infinity. If the liner chrge density is 0 x 0 /x, determine the electric field t the origin. 27. A uniformly chrged ring of rdius 10.0 cm hs totl chrge of 75.0 C. Find the electric field on the xis of the ring t () 1.00 cm, (b) 5.00 cm, (c) 30.0 cm, nd (d) 100 cm from the center of the ring. 28. Show tht the mximum field strength E mx long the xis of uniformly chrged ring occurs t x /!2 (see Fig ) nd hs the vlue Q /(6!30 2 ). 29. A uniformly chrged disk of rdius 35.0 cm crries chrge density of C/m 2. Clculte the electric field on the xis of the disk t () 5.00 cm, (b) 10.0 cm, (c) 50.0 cm, nd (d) 200 cm from the center of the disk. 30. Exmple 23.9 derives the exct expression for the electric field t point on the xis of uniformly chrged disk. Consider disk of rdius R 3.00 cm hving uniformly distributed chrge of 5.20 C. () Using the result of Exmple 23.9, compute the electric field t point on the xis nd 3.00 mm from the center. Compre this nswer with the field computed from the nerfield pproximtion E /20. (b) Using the result of Exmple 23.9, compute the electric field t point on the xis nd 30.0 cm from the center of the disk. Compre this result with the electric field obtined by treting the disk s 5.20-C point chrge t distnce of 30.0 cm. 31. The electric field long the xis of uniformly chrged disk of rdius R nd totl chrge Q ws clculted in Exmple Show tht the electric field t distnces x tht re gret compred with R pproches tht of point chrge Q R 2. (Hint: First show tht x/(x 2 R 2 ) 1/2 (1 R 2 /x 2 ) 1/2, nd use the binomil expnsion (1 ) n 1 n when 32. A piece of Styrofom hving mss m crries net chrge of q nd flots bove the center of very lrge horizontl sheet of plstic tht hs uniform chrge density on its surfce. Wht is the chrge per unit re on the plstic sheet? 33. A uniformly chrged insulting rod of length 14.0 cm is bent into the shpe of semicircle, s shown in Figure P The rod hs totl chrge of 7.50 C. Find the mgnitude nd direction of the electric field t O, the center of the semicircle. 34. () Consider uniformly chrged right circulr cylindricl shell hving totl chrge Q, rdius R, nd height h. Determine the electric field t point distnce d from the right side of the cylinder, s shown in Figure P (Hint: Use the result of Exmple 23.8 nd tret the cylinder s collection of ring chrges.) (b) Consider now solid cylinder with the sme dimensions nd V 1.) crrying the sme chrge, which is uniformly distributed through its volume. Use the result of Exmple 23.9 to find the field it cretes t the sme point. 35. A thin rod of length nd uniform chrge per unit length lies long the x xis, s shown in Figure P () Show tht the electric field t P, distnce y from the rod, long the perpendiculr bisector hs no x component nd is given by E 2k e sin 0/y. (b) Using your result to prt (), show tht the field of rod of infinite length is E 2k e/y. (Hint: First clculte the field t P due to n element of length dx, which hs chrge dx. Then chnge vribles from x to, using the fcts tht x y tn nd dx y sec 2 d, nd integrte over.) 36. Three solid plstic cylinders ll hve rdius of 2.50 cm nd length of 6.00 cm. One () crries chrge with O Figure P23.33 h dx R Figure P23.34 θ 0 y y O P θ dx Figure P23.35 d x

30 Problems 737 uniform density 15.0 nc/m 2 everywhere on its surfce. Another (b) crries chrge with the sme uniform density on its curved lterl surfce only. The third (c) crries chrge with uniform density 500 nc/m 3 throughout the plstic. Find the chrge of ech cylinder. 37. Eight solid plstic cubes, ech 3.00 cm on ech edge, re glued together to form ech one of the objects (i, ii, iii, nd iv) shown in Figure P () If ech object crries chrge with uniform density of 400 nc/m 3 throughout its volume, wht is the chrge of ech object? (b) If ech object is given chrge with uniform density of 15.0 nc/m 2 everywhere on its exposed surfce, wht is the chrge on ech object? (c) If chrge is plced only on the edges where perpendiculr surfces meet, with uniform density of 80.0 pc/m, wht is the chrge of ech object? (i) (ii) (iii) (iv) Section 23.6 Figure P23.37 Electric Field Lines 38. A positively chrged disk hs uniform chrge per unit re s described in Exmple Sketch the electric field lines in plne perpendiculr to the plne of the disk pssing through its center. 39. A negtively chrged rod of finite length hs uniform chrge per unit length. Sketch the electric field lines in plne contining the rod. 40. Figure P23.40 shows the electric field lines for two point chrges seprted by smll distnce. () Determine the rtio q 1 /q 2. (b) Wht re the signs of q 1 nd q 2? q 2 q 1 Figure P23.40 WEB Section 23.7 Motion of Chrged Prticles in Uniform Electric Field 41. An electron nd proton re ech plced t rest in n electric field of 520 N/C. Clculte the speed of ech prticle 48.0 ns fter being relesed. 42. A proton is projected in the positive x direction into region of uniform electric field E i N/C. The proton trvels 7.00 cm before coming to rest. Determine () the ccelertion of the proton, (b) its initil speed, nd (c) the time it tkes the proton to come to rest. 43. A proton ccelertes from rest in uniform electric field of 640 N/C. At some lter time, its speed hs reched m/s (nonreltivistic, since v is much less thn the speed of light). () Find the ccelertion of the proton. (b) How long does it tke the proton to rech this speed? (c) How fr hs it moved in this time? (d) Wht is its kinetic energy t this time? 44. The electrons in prticle bem ech hve kinetic energy of J. Wht re the mgnitude nd direction of the electric field tht stops these electrons in distnce of 10.0 cm? 45. The electrons in prticle bem ech hve kinetic energy K. Wht re the mgnitude nd direction of the electric field tht stops these electrons in distnce d? 46. A positively chrged bed hving mss of 1.00 g flls from rest in vcuum from height of 5.00 m in uniform verticl electric field with mgnitude of N/C. The bed hits the ground t speed of 21.0 m/s. Determine () the direction of the electric field (up or down) nd (b) the chrge on the bed. 47. A proton moves t m/s in the horizontl direction. It enters uniform verticl electric field with mgnitude of N/C. Ignoring ny grvittionl effects, find () the time it tkes the proton to trvel 5.00 cm horizontlly, (b) its verticl displcement fter it hs trveled 5.00 cm horizontlly, nd (c) the horizontl nd verticl components of its velocity fter it hs trveled 5.00 cm horizontlly. 48. An electron is projected t n ngle of 30.0 bove the horizontl t speed of m/s in region where the electric field is E 390j N/C. Neglecting the effects of grvity, find () the time it tkes the electron to return to its initil height, (b) the mximum height it reches, nd (c) its horizontl displcement when it reches its mximum height. 49. Protons re projected with n initil speed v i m/s into region where uniform electric field E (720j) N/C is present, s shown in Figure P The protons re to hit trget tht lies t horizontl distnce of 1.27 mm from the point where the protons re lunched. Find () the two projection ngles tht result in hit nd (b) the totl time of flight for ech trjectory.

31 738 CHAPTER 23 Electric Fields Proton bem ADDITIONAL PROBLEMS v i θ 1.27 mm Trget Figure P23.49 E = (720 j) N/C 50. Three point chrges re ligned long the x xis s shown in Figure P Find the electric field t () the position (2.00, 0) nd (b) the position (0, 2.00). WEB mkes 15.0 ngle with the verticl, wht is the net chrge on the bll? 53. A chrged cork bll of mss 1.00 g is suspended on light string in the presence of uniform electric field, s shown in Figure P When E (3.00i 5.00j) 10 5 N/C, the bll is in equilibrium t Find () the chrge on the bll nd (b) the tension in the string. 54. A chrged cork bll of mss m is suspended on light string in the presence of uniform electric field, s shown in Figure P When E (Ai Bj) N/C, where A nd B re positive numbers, the bll is in equilibrium t the ngle. Find () the chrge on the bll nd (b) the tension in the string m y m y θ E 4.00 nc 5.00 nc 3.00 nc x x Figure P A uniform electric field of mgnitude 640 N/C exists between two prllel pltes tht re 4.00 cm prt. A proton is relesed from the positive plte t the sme instnt tht n electron is relesed from the negtive plte. () Determine the distnce from the positive plte t which the two pss ech other. (Ignore the electricl ttrction between the proton nd electron.) (b) Repet prt () for sodium ion (N ) nd chlorine ion (Cl ). 52. A smll, 2.00-g plstic bll is suspended by 20.0-cmlong string in uniform electric field, s shown in Figure P If the bll is in equilibrium when the string Figure P23.53 Problems 53 nd Four identicl point chrges (q 10.0 C) re locted on the corners of rectngle, s shown in Figure P The dimensions of the rectngle re L 60.0 cm nd W 15.0 cm. Clculte the mgnitude nd direction of the net electric force exerted on the chrge t the lower left corner by the other three chrges. q y q q y E = i N/C W x 20.0 cm q L q x 15.0 Figure P23.55 m = 2.00 g Figure P Three identicl smll Styrofom blls (m 2.00 g) re suspended from fixed point by three nonconducting threds, ech with length of 50.0 cm nd with negligi-

32 Problems 739 ble mss. At equilibrium the three blls form n equilterl tringle with sides of 30.0 cm. Wht is the common chrge q crried by ech bll? 57. Two identicl metllic blocks resting on frictionless horizontl surfce re connected by light metllic spring hving the spring constnt k 100 N/m nd n unstretched length of m, s shown in Figure P A totl chrge of Q is slowly plced on the system, cusing the spring to stretch to n equilibrium length of m, s shown in Figure P23.57b. Determine the vlue of Q, ssuming tht ll the chrge resides on the blocks nd tht the blocks re like point chrges. 58. Two identicl metllic blocks resting on frictionless horizontl surfce re connected by light metllic spring hving spring constnt k nd n unstretched length L i, s shown in Figure P A totl chrge of Q is slowly plced on the system, cusing the spring to stretch to n equilibrium length L, s shown in Figure P23.57b. Determine the vlue of Q, ssuming tht ll the chrge resides on the blocks nd tht the blocks re like point chrges. y 1 N/C. Will the chrged prticle remin nonreltivistic for shorter or longer time in much lrger electric field? 61. A line of positive chrge is formed into semicircle of rdius R 60.0 cm, s shown in Figure P The chrge per unit length long the semicircle is described by the expression 0 cos. The totl chrge on the semicircle is 12.0 C. Clculte the totl force on chrge of 3.00 C plced t the center of curvture. b b b Figure P23.59 y x m k m θ R () x m k m Figure P23.61 Figure P23.57 Problems 57 nd Identicl thin rods of length 2 crry equl chrges, Q, uniformly distributed long their lengths. The rods lie long the x xis with their centers seprted by distnce of b 2 (Fig. P23.59). Show tht the mgnitude of the force exerted by the left rod on the right one is given by F k e Q ln b 2 2 b A prticle is sid to be nonreltivistic s long s its speed is less thn one-tenth the speed of light, or less thn m/s. () How long will n electron remin nonreltivistic if it strts from rest in region of n electric field of 1.00 N/C? (b) How long will proton remin nonreltivistic in the sme electric field? (c) Electric fields re commonly much lrger thn (b) 62. Two smll spheres, ech of mss 2.00 g, re suspended by light strings 10.0 cm in length (Fig. P23.62). A uniform electric field is pplied in the x direction. The spheres hve chrges equl to C nd C. Determine the electric field tht enbles the spheres to be in equilibrium t n ngle of θ θ Figure P23.62 E

33 740 CHAPTER 23 Electric Fields 63. Two smll spheres of mss m re suspended from strings of length tht re connected t common point. One sphere hs chrge Q ; the other hs chrge 2Q. Assume tht the ngles 1 nd 2 tht the strings mke with the verticl re smll. () How re 1 nd 2 relted? (b) Show tht the distnce r between the spheres is r 4k e Q 2 mg 1/3 64. Three chrges of equl mgnitude q re fixed in position t the vertices of n equilterl tringle (Fig. P23.64). A fourth chrge Q is free to move long the positive x xis under the influence of the forces exerted by the three fixed chrges. Find vlue for s for which Q is in equilibrium. You will need to solve trnscendentl eqution. q y Figure P23.64 q q θ r s Q x L Figure P Review Problem. A 1.00-g cork bll with chrge of 2.00 C is suspended verticlly on m-long light string in the presence of uniform, downwrd-directed electric field of mgnitude E N/C. If the bll is displced slightly from the verticl, it oscilltes like simple pendulum. () Determine the period of this oscilltion. (b) Should grvity be included in the clcultion for prt ()? Explin. 67. Three chrges of equl mgnitude q reside t the corners of n equilterl tringle of side length (Fig. P23.67). () Find the mgnitude nd direction of the electric field t point P, midwy between the negtive chrges, in terms of k e, q, nd. (b) Where must 4q chrge be plced so tht ny chrge locted t P experiences no net electric force? In prt (b), let P be the origin nd let the distnce between the q chrge nd P be 1.00 m. Q q q L z z q q 65. Review Problem. Four identicl point chrges, ech hving chrge q, re fixed t the corners of squre of side L. A fifth point chrge Q lies distnce z long the line perpendiculr to the plne of the squre nd pssing through the center of the squre (Fig. P23.65). () Show tht the force exerted on Q by the other four chrges is F 4k e qq z z2 L 2 2 3/2 Note tht this force is directed towrd the center of the squre whether z is positive ( Q bove the squre) or negtive (Q below the squre). (b) If z is smll compred with L, the bove expression reduces to F (constnt) zk. Why does this imply tht the motion of Q is simple hrmonic, nd wht would be the period of this motion if the mss of Q were m? k q Figure P Two identicl beds ech hve mss m nd chrge q. When plced in hemisphericl bowl of rdius R with frictionless, nonconducting wlls, the beds move, nd t equilibrium they re distnce R prt (Fig. P23.68). Determine the chrge on ech bed. q /2 /2 P q

34 Problems 741 y R R P m R m y Figure P Eight point chrges, ech of mgnitude q, re locted on the corners of cube of side s, s shown in Figure P () Determine the x, y, nd z components of the resultnt force exerted on the chrge locted t point A by the other chrges. (b) Wht re the mgnitude nd direction of this resultnt force? x q s q s s q q z Point A Figure P23.69 Problems 69 nd Consider the chrge distribution shown in Figure P () Show tht the mgnitude of the electric field t the center of ny fce of the cube hs vlue of 2.18k e q /s 2. (b) Wht is the direction of the electric field t the center of the top fce of the cube? 71. A line of chrge with uniform density of 35.0 nc/m lies long the line y 15.0 cm, between the points with coordintes x 0 nd x 40.0 cm. Find the electric field it cretes t the origin. 72. Three point chrges q, 2q, nd q re locted long the x xis, s shown in Figure P Show tht the electric field t P (y W ) long the y xis is E k 3q 2 e y 4 j q q q q y q Figure P23.72 This chrge distribution, which is essentilly tht of two electric dipoles, is clled n electric qudrupole. Note tht E vries s r 4 for the qudrupole, compred with vritions of r 3 for the dipole nd r 2 for the monopole ( single chrge). 73. Review Problem. A negtively chrged prticle q is plced t the center of uniformly chrged ring, where the ring hs totl positive chrge Q,s shown in Exmple The prticle, confined to move long the x xis, is displced smll distnce x long the xis (where x V ) nd relesed. Show tht the prticle oscilltes with simple hrmonic motion with frequency f 1 2 2q k eqq m 3 1/2 74. Review Problem. An electric dipole in uniform electric field is displced slightly from its equilibrium position, s shown in Figure P23.74, where is smll nd the chrges re seprted by distnce 2. The moment of inerti of the dipole is I. If the dipole is relesed from this position, show tht its ngulr orienttion exhibits simple hrmonic motion with frequency f 1 I q 2! 2qE 2 Figure P23.74 q q θ x E

35 742 CHAPTER 23 Electric Fields ANSWERS TO QUICK QUIZZES 23.1 (b). The mount of chrge present fter rubbing is the sme s tht before; it is just distributed differently (d). Object A might be negtively chrged, but it lso might be electriclly neutrl with n induced chrge seprtion, s shown in the following figure: B A 23.3 (b). From Newton s third lw, the electric force exerted by object B on object A is equl in mgnitude to the force exerted by object A on object B nd in the opposite direction tht is, F AB F BA Nothing, if we ssume tht the source chrge producing the field is not disturbed by our ctions. Remember tht the electric field is creted not by the 3-C chrge or by the 3-C chrge but by the source chrge (unseen in this cse) A, B, nd C. The field is gretest t point A becuse this is where the field lines re closest together. The bsence of lines t point C indictes tht the electric field there is zero.

36

37 2.2 This is the Nerest One Hed 743 P U Z Z L E R Some rilwy compnies re plnning to cot the windows of their commuter trins with very thin lyer of metl. (The coting is so thin you cn see through it.) They re doing this in response to rider complints bout other pssengers tlking loudly on cellulr telephones. How cn metllic coting tht is only few hundred nnometers thick overcome this problem? (Arthur Tilley/FPG Interntionl) c h p t e r Guss s Lw Chpter Outline 24.1 Electric Flux 24.2 Guss s Lw 24.3 Appliction of Guss s Lw to Chrged Insultors 24.4 Conductors in Electrosttic Equilibrium 24.5 (Optionl) Experimentl Verifiction of Guss s Lw nd Coulomb s Lw 24.6 (Optionl) Forml Derivtion of Guss s Lw 743

38 744 CHAPTER 24 Guss s Lw Are = A E In the preceding chpter we showed how to use Coulomb s lw to clculte the electric field generted by given chrge distribution. In this chpter, we describe Guss s lw nd n lterntive procedure for clculting electric fields. The lw is bsed on the fct tht the fundmentl electrosttic force between point chrges exhibits n inverse-squre behvior. Although consequence of Coulomb s lw, Guss s lw is more convenient for clculting the electric fields of highly symmetric chrge distributions nd mkes possible useful qulittive resoning when we re deling with complicted problems. Figure 24.1 Field lines representing uniform electric field 11.6 penetrting plne of re A perpendiculr to the field. The electric flux E through this re is equl to EA ELECTRIC FLUX The concept of electric field lines is described qulittively in Chpter 23. We now use the concept of electric flux to tret electric field lines in more quntittive wy. Consider n electric field tht is uniform in both mgnitude nd direction, s shown in Figure The field lines penetrte rectngulr surfce of re A, which is perpendiculr to the field. Recll from Section 23.6 tht the number of lines per unit re (in other words, the line density) is proportionl to the mgnitude of the electric field. Therefore, the totl number of lines penetrting the surfce is proportionl to the product EA. This product of the mgnitude of the electric field E nd surfce re A perpendiculr to the field is clled the electric flux E (uppercse Greek phi): E EA (24.1) From the SI units of E nd A, we see tht E hs units of newtonmeters squred per coulomb (Nm 2 /C). Electric flux is proportionl to the number of electric field lines penetrting some surfce. EXAMPLE 24.1 Flux Through Sphere Wht is the electric flux through sphere tht hs rdius of 1.00 m nd crries chrge of 1.00 C t its center? Solution The mgnitude of the electric field 1.00 m from this chrge is given by Eqution 23.4, E k e q r 2 ( Nm 2 /C 2 ) C (1.00 m) N/C The field points rdilly outwrd nd is therefore everywhere perpendiculr to the surfce of the sphere. The flux through the sphere (whose surfce re A 4r m 2 ) is thus Exercise E EA ( N/C)(12.6 m 2 ) Nm 2 /C Wht would be the () electric field nd (b) flux through the sphere if it hd rdius of m? Answer () N/C; (b) Nm 2 /C. If the surfce under considertion is not perpendiculr to the field, the flux through it must be less thn tht given by Eqution We cn understnd this by considering Figure 24.2, in which the norml to the surfce of re A is t n ngle to the uniform electric field. Note tht the number of lines tht cross this re A is equl to the number tht cross the re A, which is projection of re A ligned perpendiculr to the field. From Figure 24.2 we see tht the two res re relted by A Acos. Becuse the flux through A equls the flux through A, we

39 24.1 Electric Flux 745 A = A cos θ θ A Norml θ E Figure 24.2 Field lines representing uniform electric field penetrting n re A tht is t n ngle to the field. Becuse the number of lines tht go through the re A is the sme s the number tht go through A, the flux through A is equl to the flux through A nd is given by E EA cos. QuickLb Shine desk lmp onto plying crd nd notice how the size of the shdow on your desk depends on the orienttion of the crd with respect to the bem of light. Could formul like Eqution 24.2 be used to describe how much light ws being blocked by the crd? conclude tht the flux through A is E EA EA cos (24.2) From this result, we see tht the flux through surfce of fixed re A hs mximum vlue EA when the surfce is perpendiculr to the field (in other words, when the norml to the surfce is prllel to the field, tht is, in Figure 24.2); the flux is zero when the surfce is prllel to the field (in other words, when the norml to the surfce is perpendiculr to the field, tht is, We ssumed uniform electric field in the preceding discussion. In more generl situtions, the electric field my vry over surfce. Therefore, our definition of flux given by Eqution 24.2 hs mening only over smll element of re. Consider generl surfce divided up into lrge number of smll elements, ech of re A. The vrition in the electric field over one element cn be neglected if the element is sufficiently smll. It is convenient to define vector A i whose mgnitude represents the re of the ith element of the surfce nd whose direction is defined to be perpendiculr to the surfce element, s shown in Figure The electric flux E through this element is where we hve used the definition of the sclr product of two vectors (A B AB cos ). By summing the contributions of ll elements, we obtin the totl flux through the surfce. 1 If we let the re of ech element pproch zero, then the number of elements pproches infinity nd the sum is replced by n integrl. Therefore, the generl definition of electric flux is E E E i A i cos E i A i lim E i A i A i :0 surfce E da 0 90). (24.3) Eqution 24.3 is surfce integrl, which mens it must be evluted over the surfce in question. In generl, the vlue of E depends both on the field pttern nd on the surfce. We re often interested in evluting the flux through closed surfce, which is defined s one tht divides spce into n inside nd n outside region, so tht one cnnot move from one region to the other without crossing the surfce. The surfce of sphere, for exmple, is closed surfce. Consider the closed surfce in Figure The vectors A i point in different directions for the vrious surfce elements, but t ech point they re norml to Figure 24.3 A i θ Definition of electric flux E i A smll element of surfce re A i. The electric field mkes n ngle with the vector A i, defined s being norml to the surfce element, nd the flux through the element is equl to E i A i cos. 1 It is importnt to note tht drwings with field lines hve their inccurcies becuse smll re element (depending on its loction) my hppen to hve too mny or too few field lines penetrting it. We stress tht the bsic definition of electric flux is E da. The use of lines is only n id for visulizing the concept.

40 746 CHAPTER 24 Guss s Lw A i θ E A i E A i θ E Figure 24.4 A closed surfce in n electric field. The re vectors A i re, by convention, norml to the surfce nd point outwrd. The flux through n re element cn be positive (element ), zero (element ), or negtive (element ). Krl Friedrich Guss Germn mthemticin nd stronomer ( ) the surfce nd, by convention, lwys point outwrd. At the element lbeled, the field lines re crossing the surfce from the inside to the outside nd hence, the flux E E Ai through this element is positive. For element, the field lines grze the surfce (perpendiculr to the vector A i ); thus, nd the flux is zero. For elements such s, where the field lines re crossing the surfce from outside to inside, 180 nd the flux is negtive becuse cos is negtive. The net flux through the surfce is proportionl to the net number of lines leving the surfce, where the net number mens the number leving the surfce minus the number entering the surfce. If more lines re leving thn entering, the net flux is positive. If more lines re entering thn leving, the net flux is negtive. Using the symbol to represent n integrl over closed surfce, we cn write the net flux E through closed surfce s 90 E E da E n da 90; 90 (24.4) where E n represents the component of the electric field norml to the surfce. Evluting the net flux through closed surfce cn be very cumbersome. However, if the field is norml to the surfce t ech point nd constnt in mgnitude, the clcultion is strightforwrd, s it ws in Exmple The next exmple lso illustrtes this point. EXAMPLE 24.2 Flux Through Cube Consider uniform electric field E oriented in the x direction. Find the net electric flux through the surfce of cube of edges, oriented s shown in Figure Solution The net flux is the sum of the fluxes through ll fces of the cube. First, note tht the flux through four of the fces (,, nd the unnumbered ones) is zero becuse E is perpendiculr to da on these fces. The net flux through fces nd is E 1 E da E da 2

41 24.2 Guss s Lw 747 z Figure 24.5 da 1 y da 3 da 4 da 2 A closed surfce in the shpe of cube in uniform electric field oriented prllel to the x xis. The net flux through the closed surfce is zero. Side is the bottom of the cube, nd side is opposite side. x E For, E is constnt nd directed inwrd but da 1 is directed outwrd thus, the flux through this fce is becuse the re of ech fce is A 2. For, E is constnt nd outwrd nd in the sme direction s da 2 ( 0 ); hence, the flux through this fce is 2 ( 180); 1 E da 1 E(cos 180)dA E 1 da EA E 2 E da E(cos 0)dA E da EA E Therefore, the net flux over ll six fces is E E 2 E GAUSS S LAW In this section we describe generl reltionship between the net electric flux through closed surfce (often clled gussin surfce) nd the chrge enclosed by the surfce. This reltionship, known s Guss s lw, is of fundmentl importnce in the study of electric fields. Let us gin consider positive point chrge q locted t the center of sphere of rdius r, s shown in Figure From Eqution 23.4 we know tht the mgnitude of the electric field everywhere on the surfce of the sphere is E k e q /r 2. As noted in Exmple 24.1, the field lines re directed rdilly outwrd nd hence perpendiculr to the surfce t every point on the surfce. Tht is, t ech surfce point, E is prllel to the vector A i representing locl element of re A i surrounding the surfce point. Therefore, E A i E A i nd from Eqution 24.4 we find tht the net flux through the gussin surfce is E E da E da E da r Figure 24.6 q Gussin surfce dai A sphericl gussin surfce of rdius r surrounding point chrge q. When the chrge is t the center of the sphere, the electric field is everywhere norml to the surfce nd constnt in mgnitude. E where we hve moved E outside of the integrl becuse, by symmetry, E is constnt over the surfce nd given by E k e q /r 2. Furthermore, becuse the surfce is sphericl, da A 4r 2. Hence, the net flux through the gussin surfce is E k eq r 2 (4r 2 ) 4k e q Reclling from Section 23.3 tht k e 1/(40), we cn write this eqution in the form E q (24.5) We cn verify tht this expression for the net flux gives the sme result s Exmple 24.1: /( C 2 /Nm 2 ) Nm 2 E ( C) /C. 0

42 748 CHAPTER 24 Guss s Lw S 3 S 2 S 1 q Figure 24.7 Closed surfces of vrious shpes surrounding chrge q. The net electric flux is the sme through ll surfces. q Figure 24.8 A point chrge locted outside closed surfce. The number of lines entering the surfce equls the number leving the surfce. The net electric flux through closed surfce is zero if there is no chrge inside Note from Eqution 24.5 tht the net flux through the sphericl surfce is proportionl to the chrge inside. The flux is independent of the rdius r becuse the re of the sphericl surfce is proportionl to r 2, wheres the electric field is proportionl to 1/r 2. Thus, in the product of re nd electric field, the dependence on r cncels. Now consider severl closed surfces surrounding chrge q, s shown in Figure Surfce S 1 is sphericl, but surfces S 2 nd S 3 re not. From Eqution 24.5, the flux tht psses through S 1 hs the vlue q/ 0. As we discussed in the previous section, flux is proportionl to the number of electric field lines pssing through surfce. The construction shown in Figure 24.7 shows tht the number of lines through S 1 is equl to the number of lines through the nonsphericl surfces S 2 nd S 3. Therefore, we conclude tht the net flux through ny closed surfce is independent of the shpe of tht surfce. The net flux through ny closed surfce surrounding point chrge q is given by q/ 0. Now consider point chrge locted outside closed surfce of rbitrry shpe, s shown in Figure As you cn see from this construction, ny electric field line tht enters the surfce leves the surfce t nother point. The number of electric field lines entering the surfce equls the number leving the surfce. Therefore, we conclude tht the net electric flux through closed surfce tht surrounds no chrge is zero. If we pply this result to Exmple 24.2, we cn esily see tht the net flux through the cube is zero becuse there is no chrge inside the cube. Quick Quiz 24.1 Suppose tht the chrge in Exmple 24.1 is just outside the sphere, 1.01 m from its center. Wht is the totl flux through the sphere? Let us extend these rguments to two generlized cses: (1) tht of mny point chrges nd (2) tht of continuous distribution of chrge. We once gin use the superposition principle, which sttes tht the electric field due to mny chrges is the vector sum of the electric fields produced by the individul chrges. Therefore, we cn express the flux through ny closed surfce s E da (E 1 E 2 ) da where E is the totl electric field t ny point on the surfce produced by the vector ddition of the electric fields t tht point due to the individul chrges.

43 24.2 Guss s Lw 749 Consider the system of chrges shown in Figure The surfce S surrounds only one chrge, q 1 ; hence, the net flux through S is q 1 / 0. The flux through S due to chrges q 2 nd q 3 outside it is zero becuse ech electric field line tht enters S t one point leves it t nother. The surfce S surrounds chrges q 2 nd q 3 ; hence, the net flux through it is (q 2 q 3 )/0. Finlly, the net flux through surfce S is zero becuse there is no chrge inside this surfce. Tht is, ll the electric field lines tht enter S t one point leve t nother. Guss s lw, which is generliztion of wht we hve just described, sttes tht the net flux through ny closed surfce is E E da q in 0 (24.6) Guss s lw where q in represents the net chrge inside the surfce nd E represents the electric field t ny point on the surfce. A forml proof of Guss s lw is presented in Section When using Eqution 24.6, you should note tht lthough the chrge q in is the net chrge inside the gussin surfce, E represents the totl electric field, which includes contributions from chrges both inside nd outside the surfce. In principle, Guss s lw cn be solved for E to determine the electric field due to system of chrges or continuous distribution of chrge. In prctice, however, this type of solution is pplicble only in limited number of highly symmetric situtions. As we shll see in the next section, Guss s lw cn be used to evlute the electric field for chrge distributions tht hve sphericl, cylindricl, or plnr symmetry. If one chooses the gussin surfce surrounding the chrge distribution crefully, the integrl in Eqution 24.6 cn be simplified. You should lso note tht gussin surfce is mthemticl construction nd need not coincide with ny rel physicl surfce. Quick Quiz 24.2 For gussin surfce through which the net flux is zero, the following four sttements could be true. Which of the sttements must be true? () There re no chrges inside the surfce. (b) The net chrge inside the surfce is zero. (c) The electric field is zero everywhere on the surfce. (d) The number of electric field lines entering the surfce equls the number leving the surfce. Guss s lw is useful for evluting E when the chrge distribution hs high symmetry S q 1 q 2 S Figure 24.9 q 3 S The net electric flux through ny closed surfce depends only on the chrge inside tht surfce. The net flux through surfce S is q 1 / 0, the net flux through surfce S is (q 2 q 3 )/0, nd the net flux through surfce S is zero. CONCEPTUAL EXAMPLE 24.3 A sphericl gussin surfce surrounds point chrge q. Describe wht hppens to the totl flux through the surfce if () the chrge is tripled, (b) the rdius of the sphere is doubled, (c) the surfce is chnged to cube, nd (d) the chrge is moved to nother loction inside the surfce. Solution () The flux through the surfce is tripled becuse flux is proportionl to the mount of chrge inside the surfce. (b) The flux does not chnge becuse ll electric field lines from the chrge pss through the sphere, regrdless of its rdius. (c) The flux does not chnge when the shpe of the gussin surfce chnges becuse ll electric field lines from the chrge pss through the surfce, regrdless of its shpe. (d) The flux does not chnge when the chrge is moved to nother loction inside tht surfce becuse Guss s lw refers to the totl chrge enclosed, regrdless of where the chrge is locted inside the surfce.

44 750 CHAPTER 24 Guss s Lw 24.3 APPLICATION OF GAUSS S LAW TO CHARGED INSULATORS As mentioned erlier, Guss s lw is useful in determining electric fields when the chrge distribution is chrcterized by high degree of symmetry. The following exmples demonstrte wys of choosing the gussin surfce over which the surfce integrl given by Eqution 24.6 cn be simplified nd the electric field determined. In choosing the surfce, we should lwys tke dvntge of the symmetry of the chrge distribution so tht we cn remove E from the integrl nd solve for it. The gol in this type of clcultion is to determine surfce tht stisfies one or more of the following conditions: 1. The vlue of the electric field cn be rgued by symmetry to be constnt over the surfce. 2. The dot product in Eqution 24.6 cn be expressed s simple lgebric product E da becuse E nd da re prllel. 3. The dot product in Eqution 24.6 is zero becuse E nd da re perpendiculr. 4. The field cn be rgued to be zero over the surfce. All four of these conditions re used in exmples throughout the reminder of this chpter. EXAMPLE 24.4 The Electric Field Due to Point Chrge Strting with Guss s lw, clculte the electric field due to n isolted point chrge q. where we hve used the fct tht the surfce re of sphere is 4r 2. Now, we solve for the electric field: Solution A single chrge represents the simplest possible chrge distribution, nd we use this fmilir cse to show how to solve for the electric field with Guss s lw. We choose sphericl gussin surfce of rdius r centered on the point chrge, s shown in Figure The electric field due to positive point chrge is directed rdilly outwrd by symmetry nd is therefore norml to the surfce t every point. Thus, s in condition (2), E is prllel to da t ech point. Therefore, E da E da nd Guss s lw gives E E da E da q By symmetry, E is constnt everywhere on the surfce, which stisfies condition (1), so it cn be removed from the integrl. Therefore, E da E da E(4r 2 ) q 0 0 Figure E q 40r 2 This is the fmilir electric field due to point chrge tht we developed from Coulomb s lw in Chpter 23. r q k e q r 2 Gussin surfce The point chrge q is t the center of the sphericl gussin surfce, nd E is prllel to d A t every point on the surfce. da E 11.6 EXAMPLE 24.5 A Sphericlly Symmetric Chrge Distribution An insulting solid sphere of rdius hs uniform volume Solution Becuse the chrge distribution is sphericlly chrge density nd crries totl positive chrge Q (Fig. symmetric, we gin select sphericl gussin surfce of rdius 24.11). () Clculte the mgnitude of the electric field t r, concentric with the sphere, s shown in Figure point outside the sphere. For this choice, conditions (1) nd (2) re stisfied, s they

45 24.3 Appliction of Guss s Lw to Chrged Insultors 751 were for the point chrge in Exmple Following the line of resoning given in Exmple 24.4, we find tht (for r ) Note tht this result is identicl to the one we obtined for point chrge. Therefore, we conclude tht, for uniformly chrged sphere, the field in the region externl to the sphere is equivlent to tht of point chrge locted t the center of the sphere. (b) Find the mgnitude of the electric field t point inside the sphere. Solution E k e Q r 2 In this cse we select sphericl gussin surfce hving rdius r, concentric with the insulted sphere (Fig b). Let us denote the volume of this smller sphere by V. To pply Guss s lw in this sitution, it is importnt to recognize tht the chrge q in within the gussin surfce of volume V is less thn Q. To clculte q in, we use the fct tht q in V : q in V ( 4 3r 3 ) By symmetry, the mgnitude of the electric field is constnt everywhere on the sphericl gussin surfce nd is norml to the surfce t ech point both conditions (1) nd (2) re stisfied. Therefore, Guss s lw in the region r gives Solving for E gives Becuse by definition nd since k e 1/(40), this expression for E cn be written s E E da E da E(4r 2 ) q in E Q / q in 40r 2 Qr 40 3 (for r ) Note tht this result for E differs from the one we obtined in prt (). It shows tht E : 0 s r : 0. Therefore, the result elimintes the problem tht would exist t r 0 if E vried s 1/r 2 inside the sphere s it does outside the sphere. Tht is, if E 1/r 2 for r, the field would be infinite t r 0, which is physiclly impossible. Note lso tht the expressions for prts () nd (b) mtch when r. A plot of E versus r is shown in Figure r 3 40r 2 k e Q 3 r 30 0 r r Figure () Gussin sphere r (b) Gussin sphere A uniformly chrged insulting sphere of rdius nd totl chrge Q. () The mgnitude of the electric field t point exterior to the sphere is k e Q /r 2. (b) The mgnitude of the electric field inside the insulting sphere is due only to the chrge within the gussin sphere defined by the dshed circle nd is k e Qr / 3. Figure E E = k e Q r 2 A plot of E versus r for uniformly chrged insulting sphere. The electric field inside the sphere (r ) vries linerly with r. The field outside the sphere (r ) is the sme s tht of point chrge Q locted t r 0. r EXAMPLE 24.6 The Electric Field Due to Thin Sphericl Shell A thin sphericl shell of rdius hs totl chrge Q distributed uniformly over its surfce (Fig ). Find the electric field t points () outside nd (b) inside the shell. the shell is equivlent to tht due to point chrge Q locted t the center: Solution () The clcultion for the field outside the shell is identicl to tht for the solid sphere shown in Exmple If we construct sphericl gussin surfce of rdius r concentric with the shell (Fig b), the chrge inside this surfce is Q. Therefore, the field t point outside E k e Q r 2 (for r ) (b) The electric field inside the sphericl shell is zero. This follows from Guss s lw pplied to sphericl surfce of rdius r concentric with the shell (Fig c). Becuse

46 752 CHAPTER 24 Guss s Lw of the sphericl symmetry of the chrge distribution nd becuse the net chrge inside the surfce is zero stisfction of conditions (1) nd (2) gin ppliction of Guss s lw shows tht E 0 in the region r. We obtin the sme results using Eqution 23.6 nd integrting over the chrge distribution. This clcultion is rther complicted. Guss s lw llows us to determine these results in much simpler wy. Gussin surfce Gussin surfce E in = 0 E r r Figure () () The electric field inside uniformly chrged sphericl shell is zero. The field outside is the sme s tht due to point chrge Q locted t the center of the shell. (b) Gussin surfce for r. (c) Gussin surfce for r. (b) (c) EXAMPLE 24.7 A Cylindriclly Symmetric Chrge Distribution 11.7 Find the electric field distnce r from line of positive chrge of infinite length nd constnt chrge per unit length (Fig ). Solution The symmetry of the chrge distribution requires tht E be perpendiculr to the line chrge nd directed outwrd, s shown in Figure nd b. To reflect the symmetry of the chrge distribution, we select cylindricl gussin surfce of rdius r nd length tht is coxil with the line chrge. For the curved prt of this surfce, E is constnt in mgnitude nd perpendiculr to the surfce t ech point stisfction of conditions (1) nd (2). Furthermore, the flux through the ends of the gussin cylinder is zero becuse E is prllel to these surfces the first ppliction we hve seen of condition (3). We tke the surfce integrl in Guss s lw over the entire gussin surfce. Becuse of the zero vlue of E da for the ends of the cylinder, however, we cn restrict our ttention to only the curved surfce of the cylinder. The totl chrge inside our gussin surfce is. Applying Guss s lw nd conditions (1) nd (2), we find tht for the curved surfce Gussin surfce () r da E E E E da E da EA q in 0 0 Figure () An infinite line of chrge surrounded by cylindricl gussin surfce concentric with the line. (b) An end view shows tht the electric field t the cylindricl surfce is constnt in mgnitude nd perpendiculr to the surfce. (b)

47 24.3 Appliction of Guss s Lw to Chrged Insultors 753 The re of the curved surfce is A 2r; therefore, E E(2r) 20r (24.7) Thus, we see tht the electric field due to cylindriclly symmetric chrge distribution vries s 1/r, wheres the field externl to sphericlly symmetric chrge distribution vries s 1/r 2. Eqution 24.7 ws lso derived in Chpter 23 (see Problem 35[b]), by integrtion of the field of point chrge. If the line chrge in this exmple were of finite length, the result for E would not be tht given by Eqution A finite line chrge does not possess sufficient symmetry for us to mke use of Guss s lw. This is becuse the mgnitude of 0 2k e r the electric field is no longer constnt over the surfce of the gussin cylinder the field ner the ends of the line would be different from tht fr from the ends. Thus, condition (1) would not be stisfied in this sitution. Furthermore, E is not perpendiculr to the cylindricl surfce t ll points the field vectors ner the ends would hve component prllel to the line. Thus, condition (2) would not be stisfied. When there is insufficient symmetry in the chrge distribution, s in this sitution, it is necessry to use Eqution 23.6 to clculte E. For points close to finite line chrge nd fr from the ends, Eqution 24.7 gives good pproximtion of the vlue of the field. It is left for you to show (see Problem 29) tht the electric field inside uniformly chrged rod of finite rdius nd infinite length is proportionl to r. EXAMPLE 24.8 A Nonconducting Plne of Chrge Find the electric field due to nonconducting, infinite plne of positive chrge with uniform surfce chrge density. Solution By symmetry, E must be perpendiculr to the plne nd must hve the sme mgnitude t ll points equidistnt from the plne. The fct tht the direction of E is wy from positive chrges indictes tht the direction of E on one side of the plne must be opposite its direction on the other side, s shown in Figure A gussin surfce tht reflects the symmetry is smll cylinder whose xis is perpendiculr to the plne nd whose ends ech hve n re A nd re equidistnt from the plne. Becuse E is prllel to the curved surfce nd, therefore, perpendiculr to da everywhere on the surfce condition (3) is stisfied nd there is no contribution to the surfce integrl from this surfce. For the flt ends of the cylinder, conditions (1) nd (2) re stisfied. The flux through ech end of the cylinder is EA; hence, the totl flux through the entire gussin surfce is just tht through the ends, E 2EA. Noting tht the totl chrge inside the surfce is q in A, we use Guss s lw nd find tht E 2EA q in E 20 0 A 0 (24.8) Becuse the distnce from ech flt end of the cylinder to the plne does not pper in Eqution 24.8, we conclude tht E /2 0 t ny distnce from the plne. Tht is, the field is uniform everywhere. An importnt chrge configurtion relted to this exmple consists of two prllel plnes, one positively chrged nd the other negtively chrged, nd ech with surfce chrge density (see Problem 58). In this sitution, the electric fields due to the two plnes dd in the region between the plnes, resulting in field of mgnitude / 0, nd cncel elsewhere to give field of zero. Figure E Gussin cylinder A cylindricl gussin surfce penetrting n infinite plne of chrge. The flux is EA through ech end of the gussin surfce nd zero through its curved surfce. E A CONCEPTUAL EXAMPLE 24.9 Explin why Guss s lw cnnot be used to clculte the electric field ner n electric dipole, chrged disk, or tringle with point chrge t ech corner.

48 754 CHAPTER 24 Guss s Lw Solution The chrge distributions of ll these configurtions do not hve sufficient symmetry to mke the use of Guss s lw prcticl. We cnnot find closed surfce surrounding ny of these distributions tht stisfies one or more of conditions (1) through (4) listed t the beginning of this section. Properties of conductor in electrosttic equilibrium E Figure A conducting slb in n externl electric field E. The chrges induced on the two surfces of the slb produce n electric field tht opposes the externl field, giving resultnt field of zero inside the slb. Figure A conductor of rbitrry shpe. The broken line represents gussin surfce just inside the conductor. E Gussin surfce 24.4 CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM As we lerned in Section 23.2, good electricl conductor contins chrges (electrons) tht re not bound to ny tom nd therefore re free to move bout within the mteril. When there is no net motion of chrge within conductor, the conductor is in electrosttic equilibrium. As we shll see, conductor in electrosttic equilibrium hs the following properties: 1. The electric field is zero everywhere inside the conductor. 2. If n isolted conductor crries chrge, the chrge resides on its surfce. 3. The electric field just outside chrged conductor is perpendiculr to the surfce of the conductor nd hs mgnitude / 0, where is the surfce chrge density t tht point. 4. On n irregulrly shped conductor, the surfce chrge density is gretest t loctions where the rdius of curvture of the surfce is smllest. We verify the first three properties in the discussion tht follows. The fourth property is presented here without further discussion so tht we hve complete list of properties for conductors in electrosttic equilibrium. We cn understnd the first property by considering conducting slb plced in n externl field E (Fig ). We cn rgue tht the electric field inside the conductor must be zero under the ssumption tht we hve electrosttic equilibrium. If the field were not zero, free chrges in the conductor would ccelerte under the ction of the field. This motion of electrons, however, would men tht the conductor is not in electrosttic equilibrium. Thus, the existence of electrosttic equilibrium is consistent only with zero field in the conductor. Let us investigte how this zero field is ccomplished. Before the externl field is pplied, free electrons re uniformly distributed throughout the conductor. When the externl field is pplied, the free electrons ccelerte to the left in Figure 24.16, cusing plne of negtive chrge to be present on the left surfce. The movement of electrons to the left results in plne of positive chrge on the right surfce. These plnes of chrge crete n dditionl electric field inside the conductor tht opposes the externl field. As the electrons move, the surfce chrge density increses until the mgnitude of the internl field equls tht of the externl field, nd the net result is net field of zero inside the conductor. The time it tkes good conductor to rech equilibrium is of the order of s, which for most purposes cn be considered instntneous. We cn use Guss s lw to verify the second property of conductor in electrosttic equilibrium. Figure shows n rbitrrily shped conductor. A gussin surfce is drwn inside the conductor nd cn be s close to the conductor s surfce s we wish. As we hve just shown, the electric field everywhere inside the conductor is zero when it is in electrosttic equilibrium. Therefore, the electric field must be zero t every point on the gussin surfce, in ccordnce with condition (4) in Section Thus, the net flux through this gussin surfce is zero. From this result nd Guss s lw, we conclude tht the net chrge inside the gussin sur-

49 24.4 Conductors in Electrosttic Equilibrium 755 Electric field pttern surrounding chrged conducting plte plced ner n oppositely chrged conducting cylinder. Smll pieces of thred suspended in oil lign with the electric field lines. Note tht (1) the field lines re perpendiculr to both conductors nd (2) there re no lines inside the cylinder (E 0). fce is zero. Becuse there cn be no net chrge inside the gussin surfce (which is rbitrrily close to the conductor s surfce), ny net chrge on the conductor must reside on its surfce. Guss s lw does not indicte how this excess chrge is distributed on the conductor s surfce. We cn lso use Guss s lw to verify the third property. We drw gussin surfce in the shpe of smll cylinder whose end fces re prllel to the surfce of the conductor (Fig ). Prt of the cylinder is just outside the conductor, nd prt is inside. The field is norml to the conductor s surfce from the condition of electrosttic equilibrium. (If E hd component prllel to the conductor s surfce, the free chrges would move long the surfce; in such cse, the conductor would not be in equilibrium.) Thus, we stisfy condition (3) in Section 24.3 for the curved prt of the cylindricl gussin surfce there is no flux through this prt of the gussin surfce becuse E is prllel to the surfce. There is no flux through the flt fce of the cylinder inside the conductor becuse here E 0 stisfction of condition (4). Hence, the net flux through the gussin surfce is tht through only the flt fce outside the conductor, where the field is perpendiculr to the gussin surfce. Using conditions (1) nd (2) for this fce, the flux is EA, where E is the electric field just outside the conductor nd A is the re of the cylinder s fce. Applying Guss s lw to this surfce, we obtin E E da EA q in where we hve used the fct tht q in A. Solving for E gives E 0 0 A 0 (24.9) E n A Figure Electric field just outside chrged conductor A gussin surfce in the shpe of smll cylinder is used to clculte the electric field just outside chrged conductor. The flux through the gussin surfce is E n A. Remember tht E is zero inside the conductor. EXAMPLE A Sphere Inside Sphericl Shell A solid conducting sphere of rdius crries net positive chrge 2Q. A conducting sphericl shell of inner rdius b nd outer rdius c is concentric with the solid sphere nd crries net chrge Q. Using Guss s lw, find the electric field in the regions lbeled,,, nd in Figure nd the chrge distribution on the shell when the entire system is in electrosttic equilibrium. Solution First note tht the chrge distributions on both the sphere nd the shell re chrcterized by sphericl symmetry round their common center. To determine the electric field t vrious distnces r from this center, we construct sphericl gussin surfce for ech of the four regions of interest. Such surfce for region is shown in Figure To find E inside the solid sphere (region ), consider

50 756 CHAPTER 24 Guss s Lw Q r 2Q b c lines must be directed rdilly outwrd nd be constnt in mgnitude on the gussin surfce. Following Exmple 24.4 nd using Guss s lw, we find tht E 2 A E 2 (4r 2 ) q in E 2 2Q 40r 2 0 2k e Q r 2 2Q 0 (for r b) Figure A solid conducting sphere of rdius nd crrying chrge 2Q surrounded by conducting sphericl shell crrying chrge Q. gussin surfce of rdius r. Becuse there cn be no chrge inside conductor in electrosttic equilibrium, we see tht q in 0; thus, on the bsis of Guss s lw nd symmetry, E 1 0 for r. In region between the surfce of the solid sphere nd the inner surfce of the shell we construct sphericl gussin surfce of rdius r where r b nd note tht the chrge inside this surfce is 2Q (the chrge on the solid sphere). Becuse of the sphericl symmetry, the electric field In region, where r c, the sphericl gussin surfce we construct surrounds totl chrge of q in 2Q (Q) Q. Therefore, ppliction of Guss s lw to this surfce gives E 4 k e Q r 2 (for r c) In region, the electric field must be zero becuse the sphericl shell is lso conductor in equilibrium. If we construct gussin surfce of rdius r where b r c, we see tht q in must be zero becuse E 3 0. From this rgument, we conclude tht the chrge on the inner surfce of the sphericl shell must be 2Q to cncel the chrge 2Q on the solid sphere. Becuse the net chrge on the shell is Q, we conclude tht its outer surfce must crry chrge Q. Quick Quiz 24.3 How would the electric flux through gussin surfce surrounding the shell in Exmple chnge if the solid sphere were off-center but still inside the shell? Optionl Section 24.5 EXPERIMENTAL VERIFICATION OF GAUSS S LAW AND COULOMB S LAW When net chrge is plced on conductor, the chrge distributes itself on the surfce in such wy tht the electric field inside the conductor is zero. Guss s lw shows tht there cn be no net chrge inside the conductor in this sitution. In this section, we investigte n experimentl verifiction of the bsence of this chrge. We hve seen tht Guss s lw is equivlent to Eqution 23.6, the expression for the electric field of distribution of chrge. Becuse this eqution rises from Coulomb s lw, we cn clim theoreticlly tht Guss s lw nd Coulomb s lw re equivlent. Hence, it is possible to test the vlidity of both lws by ttempting to detect net chrge inside conductor or, equivlently, nonzero electric field inside the conductor. If nonzero field is detected within the conductor, Guss s lw nd Coulomb s lw re invlid. Mny experiments, including

51 24.5 Experimentl Verifiction of Guss s Lw nd Coulomb s Lw 757 erly work by Frdy, Cvendish, nd Mxwell, hve been performed to detect the field inside conductor. In ll reported cses, no electric field could be detected inside conductor. Here is one of the experiments tht cn be performed. 2 A positively chrged metl bll t the end of silk thred is lowered through smll opening into n unchrged hollow conductor tht is insulted from ground (Fig ). The positively chrged bll induces negtive chrge on the inner wll of the hollow conductor, leving n equl positive chrge on the outer wll (Fig b). The presence of positive chrge on the outer wll is indicted by the deflection of the needle of n electrometer ( device used to mesure chrge nd tht mesures chrge only on the outer surfce of the conductor). The bll is then lowered nd llowed to touch the inner surfce of the hollow conductor (Fig c). Chrge is trnsferred between the bll nd the inner surfce so tht neither is chrged fter contct is mde. The needle deflection remins unchnged while this hppens, indicting tht the chrge on the outer surfce is unffected. When the bll is removed, the electrometer reding remins the sme (Fig d). Furthermore, the bll is found to be unchrged; this verifies tht chrge ws trnsferred between the bll nd the inner surfce of the hollow conductor. The overll effect is tht the chrge tht ws originlly on the bll now ppers on the hollow conductor. The fct tht the deflection of the needle on the electrometer mesuring the chrge on the outer surfce remined unchnged regrdless of wht ws hppening inside the hollow conductor indictes tht the net chrge on the system lwys resided on the outer surfce of the conductor. If we now pply nother positive chrge to the metl bll nd plce it ner the outside of the conductor, it is repelled by the conductor. This demonstrtes tht E 0 outside the conductor, finding consistent with the fct tht the conductor crries net chrge. If the chrged metl bll is now lowered into the interior of the chrged hollow conductor, it exhibits no evidence of n electric force. This shows tht E 0 inside the hollow conductor. This experiment verifies the predictions of Guss s lw nd therefore verifies Coulomb s lw. The equivlence of Guss s lw nd Coulomb s lw is due to the inverse-squre behvior of the electric force. Thus, we cn interpret this experiment s verifying the exponent of 2 in the 1/r 2 behvior of the electric force. Experiments by Willims, Fller, nd Hill in 1971 showed tht the exponent of r in Coulomb s lw is (2 ), where In the experiment we hve described, the chrged bll hnging in the hollow conductor would show no deflection even in the cse in which n externl electric field is pplied to the entire system. The field inside the conductor is still zero. This bility of conductors to block externl electric fields is utilized in mny plces, from electromgnetic shielding for computer components to thin metl cotings on the glss in irport control towers to keep rdr originting outside the tower from disrupting the electronics inside. Cellulr telephone users riding trins like the one pictured t the beginning of the chpter hve to spek loudly to be herd bove the noise of the trin. In response to complints from other pssengers, the trin compnies re considering coting the windows with thin metllic conductor. This coting, combined with the metl frme of the trin cr, blocks cellulr telephone trnsmissions into nd out of the trin. ( ) 10 16! Hollow conductor () (b) (c) (d) Figure An experiment showing tht ny chrge trnsferred to conductor resides on its surfce in electrosttic equilibrium. The hollow conductor is insulted from ground, nd the smll metl bll is supported by n insulting thred. QuickLb Wrp rdio or cordless telephone in luminum foil nd see if it still works. Does it mtter if the foil touches the ntenn? The experiment is often referred to s Frdy s ice-pil experiment becuse Frdy, the first to perform it, used n ice pil for the hollow conductor.

52 758 CHAPTER 24 Guss s Lw q Figure Ω θ A A closed surfce of rbitrry shpe surrounds point chrge q. The net electric flux through the surfce is independent of the shpe of the surfce. E Optionl Section 24.6 FORMAL DERIVATION OF GAUSS S LAW One wy of deriving Guss s lw involves solid ngles. Consider sphericl surfce of rdius r contining n re element A. The solid ngle (uppercse Greek omeg) subtended t the center of the sphere by this element is defined to be A r 2 From this eqution, we see tht hs no dimensions becuse A nd r 2 both hve dimensions L 2. The dimensionless unit of solid ngle is the sterdin. (You my wnt to compre this eqution to Eqution 10.1b, the definition of the rdin.) Becuse the surfce re of sphere is 4r 2, the totl solid ngle subtended by the sphere is 4r 2 r 2 4 sterdins Now consider point chrge q surrounded by closed surfce of rbitrry shpe (Fig ). The totl electric flux through this surfce cn be obtined by evluting E A for ech smll re element A nd summing over ll elements. The flux through ech element is E E A E A cos k e q A cos r 2 where r is the distnce from the chrge to the re element, is the ngle between the electric field E nd A for the element, nd E k e q /r 2 for point chrge. In Figure 24.22, we see tht the projection of the re element perpendiculr to the rdius vector is A cos. Thus, the quntity A cos /r 2 is equl to the solid ngle tht the surfce element A subtends t the chrge q. We lso see tht is equl to the solid ngle subtended by the re element of sphericl surfce of rdius r. Becuse the totl solid ngle t point is 4 sterdins, the totl flux A r θ θ E q Ω A cos θ A Figure The re element A subtends solid ngle (A cos )/r 2 t the chrge q.

53 Summry 759 through the closed surfce is E k e q da cos r 2 k e q d 4k e q q Thus we hve derived Guss s lw, Eqution Note tht this result is independent of the shpe of the closed surfce nd independent of the position of the chrge within the surfce. 0 SUMMARY Electric flux is proportionl to the number of electric field lines tht penetrte surfce. If the electric field is uniform nd mkes n ngle with the norml to surfce of re A, the electric flux through the surfce is E EA cos (24.2) In generl, the electric flux through surfce is E E da (24.3) surfce You need to be ble to pply Equtions 24.2 nd 24.3 in vriety of situtions, prticulrly those in which symmetry simplifies the clcultion. Guss s lw sys tht the net electric flux E through ny closed gussin surfce is equl to the net chrge inside the surfce divided by 0 : E E da q in (24.6) Using Guss s lw, you cn clculte the electric field due to vrious symmetric chrge distributions. Tble 24.1 lists some typicl results. 0 TABLE 24.1 Typicl Electric Field Clcultions Using Guss s Lw Chrge Distribution Electric Field Loction Insulting sphere of rdius R, uniform chrge density, nd totl chrge Q Thin sphericl shell of rdius R nd totl chrge Q Line chrge of infinite length nd chrge per unit length Nonconducting, infinite chrged plne hving surfce chrge density Conductor hving surfce chrge density k Q e r 2 k Q e R 3 r k e Q r 2 2k e r r R r R r R r R Outside the line Everywhere outside the plne Just outside the conductor Inside the conductor

54 760 CHAPTER 24 Guss s Lw A conductor in electrosttic equilibrium hs the following properties: 1. The electric field is zero everywhere inside the conductor. 2. Any net chrge on the conductor resides entirely on its surfce. 3. The electric field just outside the conductor is perpendiculr to its surfce nd hs mgnitude / 0, where is the surfce chrge density t tht point. 4. On n irregulrly shped conductor, the surfce chrge density is gretest where the rdius of curvture of the surfce is the smllest. Problem-Solving Hints Guss s lw, s we hve seen, is very powerful in solving problems involving highly symmetric chrge distributions. In this chpter, you encountered three kinds of symmetry: plnr, cylindricl, nd sphericl. It is importnt to review Exmples 24.4 through nd to dhere to the following procedure when using Guss s lw: Select gussin surfce tht hs symmetry to mtch tht of the chrge distribution nd stisfies one or more of the conditions listed in Section For point chrges or sphericlly symmetric chrge distributions, the gussin surfce should be sphere centered on the chrge s in Exmples 24.4, 24.5, 24.6, nd For uniform line chrges or uniformly chrged cylinders, your gussin surfce should be cylindricl surfce tht is coxil with the line chrge or cylinder s in Exmple For plnes of chrge, useful choice is cylindricl gussin surfce tht strddles the plne, s shown in Exmple These choices enble you to simplify the surfce integrl tht ppers in Guss s lw nd represents the totl electric flux through tht surfce. Evlute the q in / 0 term in Guss s lw, which mounts to clculting the totl electric chrge q in inside the gussin surfce. If the chrge density is uniform (tht is, if,, or is constnt), simply multiply tht chrge density by the length, re, or volume enclosed by the gussin surfce. If the chrge distribution is nonuniform, integrte the chrge density over the region enclosed by the gussin surfce. For exmple, if the chrge is distributed long line, integrte the expression dq where dq is the chrge on n infinitesiml length element dx. For plne of chrge, integrte dq da, where da is n infinitesiml element of re. For volume of chrge, integrte dq where dv is n infinitesiml element of volume. Once the terms in Guss s lw hve been evluted, solve for the electric field on the gussin surfce if the chrge distribution is given in the problem. Conversely, if the electric field is known, clculte the chrge distribution tht produces the field. dv, dx, QUESTIONS 1. The Sun is lower in the sky during the winter thn it is in the summer. How does this chnge the flux of sunlight hitting given re on the surfce of the Erth? How does this ffect the wether? 2. If the electric field in region of spce is zero, cn you conclude no electric chrges re in tht region? Explin. 3. If more electric field lines re leving gussin surfce thn entering, wht cn you conclude bout the net chrge enclosed by tht surfce? 4. A uniform electric field exists in region of spce in which there re no chrges. Wht cn you conclude bout the net electric flux through gussin surfce plced in this region of spce?

55 Problems If the totl chrge inside closed surfce is known but the distribution of the chrge is unspecified, cn you use Guss s lw to find the electric field? Explin. 6. Explin why the electric flux through closed surfce with given enclosed chrge is independent of the size or shpe of the surfce. 7. Consider the electric field due to nonconducting infinite plne hving uniform chrge density. Explin why the electric field does not depend on the distnce from the plne in terms of the spcing of the electric field lines. 8. Use Guss s lw to explin why electric field lines must begin or end on electric chrges. (Hint: Chnge the size of the gussin surfce.) 9. On the bsis of the repulsive nture of the force between like chrges nd the freedom of motion of chrge within the conductor, explin why excess chrge on n isolted conductor must reside on its surfce. 10. A person is plced in lrge, hollow metllic sphere tht is insulted from ground. If lrge chrge is plced on the sphere, will the person be hrmed upon touching the inside of the sphere? Explin wht will hppen if the person lso hs n initil chrge whose sign is opposite tht of the chrge on the sphere. 11. How would the observtions described in Figure differ if the hollow conductor were grounded? How would they differ if the smll chrged bll were n insultor rther thn conductor? 12. Wht other experiment might be performed on the bll in Figure to show tht its chrge ws trnsferred to the hollow conductor? 13. Wht would hppen to the electrometer reding if the chrged bll in Figure touched the inner wll of the conductor? the outer wll? 14. You my hve herd tht one of the sfer plces to be during lightning storm is inside cr. Why would this be the cse? 15. Two solid spheres, both of rdius R, crry identicl totl chrges Q. One sphere is good conductor, while the other is n insultor. If the chrge on the insulting sphere is uniformly distributed throughout its interior volume, how do the electric fields outside these two spheres compre? Are the fields identicl inside the two spheres? PROBLEMS 1, 2, 3 = strightforwrd, intermedite, chllenging = full solution vilble in the Student Solutions Mnul nd Study Guide WEB = solution posted t = Computer useful in solving problem = Interctive Physics = pired numericl/symbolic problems Section 24.1 Electric Flux 1. An electric field with mgnitude of 3.50 kn/c is pplied long the x xis. Clculte the electric flux through rectngulr plne m wide nd m long if () the plne is prllel to the yz plne; (b) the plne is prllel to the xy plne; nd (c) the plne contins the y xis, nd its norml mkes n ngle of 40.0 with the x xis. 2. A verticl electric field of mgnitude N/C exists bove the Erth s surfce on dy when thunderstorm is brewing. A cr with rectngulr size of pproximtely 6.00 m by 3.00 m is trveling long rodwy sloping downwrd t Determine the electric flux through the bottom of the cr. 3. A 40.0-cm-dimeter loop is rotted in uniform electric field until the position of mximum electric flux is found. The flux in this position is mesured to be N m 2 /C. Wht is the mgnitude of the electric field? 4. A sphericl shell is plced in uniform electric field. Find the totl electric flux through the shell. 5. Consider closed tringulr box resting within horizontl electric field of mgnitude E N/C, s shown in Figure P24.5. Clculte the electric flux through () the verticl rectngulr surfce, (b) the slnted surfce, nd (c) the entire surfce of the box cm 30.0 cm A uniform electric field i bj intersects surfce of re A. Wht is the flux through this re if the surfce lies () in the yz plne? (b) in the xz plne? (c) in the xy plne? 7. A point chrge q is locted t the center of uniform ring hving liner chrge density nd rdius, s shown in Figure P24.7. Determine the totl electric flux λ Figure P24.5 q R Figure P24.7 E

56 762 CHAPTER 24 Guss s Lw through sphere centered t the point chrge nd hving rdius R, where R. 8. A pyrmid with 6.00-m-squre bse nd height of 4.00 m is plced in verticl electric field of 52.0 N/C. Clculte the totl electric flux through the pyrmid s four slnted surfces. 9. A cone with bse rdius R nd height h is locted on horizontl tble. A horizontl uniform field E penetrtes the cone, s shown in Figure P24.9. Determine the electric flux tht enters the left-hnd side of the cone. locted very smll distnce from the center of very lrge squre on the line perpendiculr to the squre nd going through its center. Determine the pproximte electric flux through the squre due to the point chrge. (c) Explin why the nswers to prts () nd (b) re identicl. 14. Clculte the totl electric flux through the prboloidl surfce due to constnt electric field of mgnitude E 0 in the direction shown in Figure P r E Section 24.2 Guss s Lw 10. The electric field everywhere on the surfce of thin sphericl shell of rdius m is mesured to be equl to 890 N/C nd points rdilly towrd the center of the sphere. () Wht is the net chrge within the sphere s surfce? (b) Wht cn you conclude bout the nture nd distribution of the chrge inside the sphericl shell? 11. The following chrges re locted inside submrine: 5.00 C, 9.00 C, 27.0 C, nd 84.0 C. () Clculte the net electric flux through the submrine. (b) Is the number of electric field lines leving the submrine greter thn, equl to, or less thn the number entering it? 12. Four closed surfces, S 1 through S 4, together with the chrges 2Q, Q, nd Q re sketched in Figure P Find the electric flux through ech surfce. R h Figure P24.9 WEB E A point chrge Q is locted just bove the center of the flt fce of hemisphere of rdius R, s shown in Figure P Wht is the electric flux () through the curved surfce nd (b) through the flt fce? δ Figure P d Q R Figure P24.15 S 4 2Q Figure P () A point chrge q is locted distnce d from n infinite plne. Determine the electric flux through the plne due to the point chrge. (b) A point chrge q is Q Q S 2 S 1 S A point chrge of 12.0 C is plced t the center of sphericl shell of rdius 22.0 cm. Wht is the totl electric flux through () the surfce of the shell nd (b) ny hemisphericl surfce of the shell? (c) Do the results depend on the rdius? Explin. 17. A point chrge of C is inside pyrmid. Determine the totl electric flux through the surfce of the pyrmid. 18. An infinitely long line chrge hving uniform chrge per unit length lies distnce d from point O, s shown in Figure P Determine the totl electric flux through the surfce of sphere of rdius R centered t O resulting from this line chrge. (Hint: Consider both cses: when R d, nd when R d.)

57 Problems 763 λ 19. A point chrge Q 5.00 C is locted t the center of cube of side L m. In ddition, six other identicl point chrges hving q 1.00 C re positioned symmetriclly round Q, s shown in Figure P Determine the electric flux through one fce of the cube. 20. A point chrge Q is locted t the center of cube of side L. In ddition, six other identicl negtive point chrges re positioned symmetriclly round Q, s shown in Figure P Determine the electric flux through one fce of the cube. d Figure P24.18 O R 23. A chrge of 170 C is t the center of cube of side 80.0 cm. () Find the totl flux through ech fce of the cube. (b) Find the flux through the whole surfce of the cube. (c) Would your nswers to prts () or (b) chnge if the chrge were not t the center? Explin. 24. The totl electric flux through closed surfce in the shpe of cylinder is Nm 2 /C. () Wht is the net chrge within the cylinder? (b) From the informtion given, wht cn you sy bout the chrge within the cylinder? (c) How would your nswers to prts () nd (b) chnge if the net flux were Nm 2 /C? 25. The line g is digonl of cube (Fig. P24.25). A point chrge q is locted on the extension of line g, very close to vertex of the cube. Determine the electric flux through ech of the sides of the cube tht meet t the point. q d b c h g L L q q L Figure P24.19 Problems 19 nd Consider n infinitely long line chrge hving uniform chrge per unit length. Determine the totl electric flux through closed right circulr cylinder of length L nd rdius R tht is prllel to the line chrge, if the distnce between the xis of the cylinder nd the line chrge is d. (Hint: Consider both cses: when R d, nd when R d.) 22. A 10.0-C chrge locted t the origin of crtesin coordinte system is surrounded by nonconducting hollow sphere of rdius 10.0 cm. A drill with rdius of 1.00 mm is ligned long the z xis, nd hole is drilled in the sphere. Clculte the electric flux through the hole. q q Q q q WEB e Figure P24.25 Section 24.3 Appliction of Guss s Lw to Chrged Insultors 26. Determine the mgnitude of the electric field t the surfce of led-208 nucleus, which contins 82 protons nd 126 neutrons. Assume tht the led nucleus hs volume 208 times tht of one proton, nd consider proton to be sphere of rdius m. 27. A solid sphere of rdius 40.0 cm hs totl positive chrge of 26.0 C uniformly distributed throughout its volume. Clculte the mgnitude of the electric field () 0 cm, (b) 10.0 cm, (c) 40.0 cm, nd (d) 60.0 cm from the center of the sphere. 28. A cylindricl shell of rdius 7.00 cm nd length 240 cm hs its chrge uniformly distributed on its curved surfce. The mgnitude of the electric field t point 19.0 cm rdilly outwrd from its xis (mesured from the midpoint of the shell) is 36.0 kn/c. Use pproximte reltionships to find () the net chrge on the shell nd (b) the electric field t point 4.00 cm from the xis, mesured rdilly outwrd from the midpoint of the shell. 29. Consider long cylindricl chrge distribution of rdius R with uniform chrge density. Find the electric field t distnce r from the xis where r R. f

58 764 CHAPTER 24 Guss s Lw 30. A nonconducting wll crries uniform chrge density of 8.60 C/cm 2. Wht is the electric field 7.00 cm in front of the wll? Does your result chnge s the distnce from the wll is vried? 31. Consider thin sphericl shell of rdius 14.0 cm with totl chrge of 32.0 C distributed uniformly on its surfce. Find the electric field () 10.0 cm nd (b) 20.0 cm from the center of the chrge distribution. 32. In nucler fission, nucleus of urnium-238, which contins 92 protons, divides into two smller spheres, ech hving 46 protons nd rdius of m. Wht is the mgnitude of the repulsive electric force pushing the two spheres prt? 33. Fill two rubber blloons with ir. Suspend both of them from the sme point on strings of equl length. Rub ech with wool or your hir, so tht they hng prt with noticeble seprtion between them. Mke order-ofmgnitude estimtes of () the force on ech, (b) the chrge on ech, (c) the field ech cretes t the center of the other, nd (d) the totl flux of electric field creted by ech blloon. In your solution, stte the quntities you tke s dt nd the vlues you mesure or estimte for them. 34. An insulting sphere is 8.00 cm in dimeter nd crries 5.70-C chrge uniformly distributed throughout its interior volume. Clculte the chrge enclosed by concentric sphericl surfce with rdius () r 2.00 cm nd (b) r 6.00 cm. 35. A uniformly chrged, stright filment 7.00 m in length hs totl positive chrge of 2.00 C. An unchrged crdbord cylinder 2.00 cm in length nd 10.0 cm in rdius surrounds the filment t its center, with the filment s the xis of the cylinder. Using resonble pproximtions, find () the electric field t the surfce of the cylinder nd (b) the totl electric flux through the cylinder. 36. The chrge per unit length on long, stright filment is 90.0 C/m. Find the electric field () 10.0 cm, (b) 20.0 cm, nd (c) 100 cm from the filment, where distnces re mesured perpendiculr to the length of the filment. 37. A lrge flt sheet of chrge hs chrge per unit re of 9.00 C/m 2. Find the electric field just bove the surfce of the sheet, mesured from its midpoint. Section 24.4 Conductors in Electrosttic Equilibrium 38. On cler, sunny dy, verticl electricl field of bout 130 N/C points down over flt ground. Wht is the surfce chrge density on the ground for these conditions? 39. A long, stright metl rod hs rdius of 5.00 cm nd chrge per unit length of 30.0 nc/m. Find the electric field () 3.00 cm, (b) 10.0 cm, nd (c) 100 cm from the xis of the rod, where distnces re mesured perpendiculr to the rod. 40. A very lrge, thin, flt plte of luminum of re A hs totl chrge Q uniformly distributed over its surfces. If the sme chrge is spred uniformly over the upper surfce of n otherwise identicl glss plte, compre the electric fields just bove the center of the upper surfce of ech plte. 41. A squre plte of copper with 50.0-cm sides hs no net chrge nd is plced in region of uniform electric field of 80.0 kn/c directed perpendiculrly to the plte. Find () the chrge density of ech fce of the plte nd (b) the totl chrge on ech fce. 42. A hollow conducting sphere is surrounded by lrger concentric, sphericl, conducting shell. The inner sphere hs chrge Q, nd the outer sphere hs chrge 3Q. The chrges re in electrosttic equilibrium. Using Guss s lw, find the chrges nd the electric fields everywhere. 43. Two identicl conducting spheres ech hving rdius of cm re connected by light 2.00-m-long conducting wire. Determine the tension in the wire if 60.0 C is plced on one of the conductors. (Hint: Assume tht the surfce distribution of chrge on ech sphere is uniform.) 44. The electric field on the surfce of n irregulrly shped conductor vries from 56.0 kn/c to 28.0 kn/c. Clculte the locl surfce chrge density t the point on the surfce where the rdius of curvture of the surfce is () gretest nd (b) smllest. 45. A long, stright wire is surrounded by hollow metl cylinder whose xis coincides with tht of the wire. The wire hs chrge per unit length of, nd the cylinder hs net chrge per unit length of 2. From this informtion, use Guss s lw to find () the chrge per unit length on the inner nd outer surfces of the cylinder nd (b) the electric field outside the cylinder, distnce r from the xis. 46. A conducting sphericl shell of rdius 15.0 cm crries net chrge of 6.40 C uniformly distributed on its surfce. Find the electric field t points () just outside the shell nd (b) inside the shell. WEB 47. A thin conducting plte 50.0 cm on side lies in the xy plne. If totl chrge of C is plced on the plte, find () the chrge density on the plte, (b) the electric field just bove the plte, nd (c) the electric field just below the plte. 48. A conducting sphericl shell hving n inner rdius of nd n outer rdius of b crries net chrge Q. If point chrge q is plced t the center of this shell, determine the surfce chrge density on () the inner surfce of the shell nd (b) the outer surfce of the shell. 49. A solid conducting sphere of rdius 2.00 cm hs chrge 8.00 C. A conducting sphericl shell of inner rdius 4.00 cm nd outer rdius 5.00 cm is concentric with the solid sphere nd hs chrge 4.00 C. Find the electric field t () r 1.00 cm, (b) r 3.00 cm, (c) r 4.50 cm, nd (d) r 7.00 cm from the center of this chrge configurtion.

59 Problems A positive point chrge is t distnce of R/2 from the center of n unchrged thin conducting sphericl shell of rdius R. Sketch the electric field lines set up by this rrngement both inside nd outside the shell. 3Q Q r (Optionl) Section 24.5 Experimentl Verifiction of Guss s Lw nd Coulomb s Lw Section 24.6 Forml Derivtion of Guss s Lw 51. A sphere of rdius R surrounds point chrge Q, locted t its center. () Show tht the electric flux through circulr cp of hlf-ngle (Fig. P24.51) is E Q 20 (1 cos ) Wht is the flux for (b) 90 nd (c) 180? ADDITIONAL PROBLEMS R Figure P A nonuniform electric field is given by the expression E yi bzj cxk, where, b, nd c re constnts. Determine the electric flux through rectngulr surfce in the xy plne, extending from x 0 to x w nd from y 0 to y h. 53. A solid insulting sphere of rdius crries net positive chrge 3Q, uniformly distributed throughout its volume. Concentric with this sphere is conducting sphericl shell with inner rdius b nd outer rdius c, nd hving net chrge Q, s shown in Figure P () Construct sphericl gussin surfce of rdius r c nd find the net chrge enclosed by this surfce. (b) Wht is the direction of the electric field t r c? (c) Find the electric field t r c. (d) Find the electric field in the region with rdius r where c r b. (e) Construct sphericl gussin surfce of rdius r, where c r b, nd find the net chrge enclosed by this surfce. (f) Construct sphericl gussin surfce of rdius r, where b r, nd find the net chrge enclosed by this surfce. (g) Find the electric field in the region b r. (h) Construct sphericl gussin surfce of rdius r, nd find n expression for the θ Q WEB Figure P24.53 net chrge enclosed by this surfce, s function of r. Note tht the chrge inside this surfce is less thn 3Q. (i) Find the electric field in the region r. (j) Determine the chrge on the inner surfce of the conducting shell. (k) Determine the chrge on the outer surfce of the conducting shell. (l) Mke plot of the mgnitude of the electric field versus r. 54. Consider two identicl conducting spheres whose surfces re seprted by smll distnce. One sphere is given lrge net positive chrge, while the other is given smll net positive chrge. It is found tht the force between them is ttrctive even though both spheres hve net chrges of the sme sign. Explin how this is possible. 55. A solid, insulting sphere of rdius hs uniform chrge density nd totl chrge Q. Concentric with this sphere is n unchrged, conducting hollow sphere whose inner nd outer rdii re b nd c, s shown in Figure P () Find the mgnitude of the electric field in the regions r, r b, b r c, nd r c. (b) Determine the induced chrge per unit re on the inner nd outer surfces of the hollow sphere. b c Figure P24.55 Problems 55 nd For the configurtion shown in Figure P24.55, suppose tht 5.00 cm, b 20.0 cm, nd c 25.0 cm. Furthermore, suppose tht the electric field t point 10.0 cm from the center is N/C rdilly inwrd, while the electric field t point 50.0 cm from the center is N/C rdilly outwrd. From this informtion, find () the chrge on the insulting sphere, b c Insultor Conductor

60 766 CHAPTER 24 Guss s Lw (b) the net chrge on the hollow conducting sphere, nd (c) the totl chrge on the inner nd outer surfces of the hollow conducting sphere. 57. An infinitely long cylindricl insulting shell of inner rdius nd outer rdius b hs uniform volume chrge density (C/m 3 ). A line of chrge density (C/m) is plced long the xis of the shell. Determine the electric field intensity everywhere. 58. Two infinite, nonconducting sheets of chrge re prllel to ech other, s shown in Figure P The sheet on the left hs uniform surfce chrge density, nd the one on the right hs uniform chrge density. Clculte the vlue of the electric field t points () to the left of, (b) in between, nd (c) to the right of the two sheets. (Hint: See Exmple 24.8.) the size of the cvity with uniform negtive chrge density.) 61. Review Problem. An erly (incorrect) model of the hydrogen tom, suggested by J. J. Thomson, proposed tht positive cloud of chrge e ws uniformly distributed throughout the volume of sphere of rdius R, with the electron n equl-mgnitude negtive point chrge e t the center. () Using Guss s lw, show tht the electron would be in equilibrium t the center nd, if displced from the center distnce r R, would experience restoring force of the form F Kr, where K is constnt. (b) Show tht K k e e 2 /R 3. (c) Find n expression for the frequency f of simple hrmonic oscilltions tht n electron of mss m e would undergo if displced short distnce ( R ) from the center nd relesed. (d) Clculte numericl vlue for R tht would result in frequency of electron vibrtion of Hz, the frequency of the light in the most intense line in the hydrogen spectrum. 62. A closed surfce with dimensions b m nd c m is locted s shown in Figure P The electric field throughout the region is nonuniform nd given by E ( x 2 )i N/C, where x is in meters. Clculte the net electric flux leving the closed surfce. Wht net chrge is enclosed by the surfce? σ σ Figure P24.58 y c E WEB 59. Repet the clcultions for Problem 58 when both sheets hve positive uniform surfce chrge densities of vlue. 60. A sphere of rdius 2 is mde of nonconducting mteril tht hs uniform volume chrge density. (Assume tht the mteril does not ffect the electric field.) A sphericl cvity of rdius is now removed from the sphere, s shown in Figure P Show tht the electric field within the cvity is uniform nd is given by E x 0 nd E y /30. (Hint: The field within the cvity is the superposition of the field due to the originl uncut sphere, plus the field due to sphere y 2 Figure P24.60 x z Figure P A solid insulting sphere of rdius R hs nonuniform chrge density tht vries with r ccording to the expression where A is constnt nd r R is mesured from the center of the sphere. () Show tht the electric field outside (r R ) the sphere is E AR 5 /50r 2. (b) Show tht the electric field inside (r R) the sphere is E Ar 3 /50. (Hint: Note tht the totl chrge Q on the sphere is equl to the integrl of dv, where r extends from 0 to R; lso note tht the chrge q within rdius r R is less thn Q. To evlute the integrls, note tht the volume element dv for sphericl shell of rdius r nd thickness dr is equl to 4r 2 dr.) 64. A point chrge Q is locted on the xis of disk of rdius R t distnce b from the plne of the disk (Fig. P24.64). Show tht if one fourth of the electric flux from the chrge psses through the disk, then R!3b. Ar 2, b x

61 Answers to Quick Quizzes 767 frequency described by the expression f 1 2! e m e0 R y b Q Figure P24.64 O x 65. A sphericlly symmetric chrge distribution hs chrge density given by where is constnt. Find the electric field s function of r. (Hint: Note tht the chrge within sphere of rdius R is equl to the integrl of dv, where r extends from 0 to R. To evlute the integrl, note tht the volume element dv for sphericl shell of rdius r nd thickness dr is equl to 4r 2 dr.) 66. An infinitely long insulting cylinder of rdius R hs volume chrge density tht vries with the rdius s /r, 0 r b where 0,, nd b re positive constnts nd r is the distnce from the xis of the cylinder. Use Guss s lw to determine the mgnitude of the electric field t rdil distnces () r R nd (b) r R. 67. Review Problem. A slb of insulting mteril (infinite in two of its three dimensions) hs uniform positive chrge density. An edge view of the slb is shown in Figure P () Show tht the mgnitude of the electric field distnce x from its center nd inside the slb is E x/0. (b) Suppose tht n electron of chrge e nd mss m e is plced inside the slb. If it is relesed from rest t distnce x from the center, show tht the electron exhibits simple hrmonic motion with d Figure P24.67 Problems 67 nd A slb of insulting mteril hs nonuniform positive chrge density where x is mesured from the center of the slb, s shown in Figure P24.67, nd C is constnt. The slb is infinite in the y nd z directions. Derive expressions for the electric field in () the exterior regions nd (b) the interior region of the slb (d/2 x d/2). 69. () Using the mthemticl similrity between Coulomb s lw nd Newton s lw of universl grvittion, show tht Guss s lw for grvittion cn be written s Cx 2, g da 4Gm in where m in is the mss inside the gussin surfce nd g F g /m represents the grvittionl field t ny point on the gussin surfce. (b) Determine the grvittionl field t distnce r from the center of the Erth where r R E, ssuming tht the Erth s mss density is uniform. ANSWERS TO QUICK QUIZZES 24.1 Zero, becuse there is no net chrge within the surfce (b) nd (d). Sttement () is not necessrily true becuse n equl number of positive nd negtive chrges could be present inside the surfce. Sttement (c) is not necessrily true, s cn be seen from Figure 24.8: A nonzero electric field exists everywhere on the surfce, but the chrge is not enclosed within the surfce; thus, the net flux is zero Any gussin surfce surrounding the system encloses the sme mount of chrge, regrdless of how the components of the system re moved. Thus, the flux through the gussin surfce would be the sme s it is when the sphere nd shell re concentric.

62

63 P U Z Z L E R Jennifer is holding on to n electriclly chrged sphere tht reches n electric potentil of bout V. The device tht genertes this high electric potentil is clled Vn de Grff genertor. Wht cuses Jennifer s hir to stnd on end like the needles of porcupine? Why is she sfe in this sitution in view of the fct tht 110 V from wll outlet cn kill you? (Henry Lep nd Jim Lehmn) c h p t e r Electric Potentil Chpter Outline 25.1 Potentil Difference nd Electric Potentil 25.2 Potentil Differences in Uniform Electric Field 25.3 Electric Potentil nd Potentil Energy Due to Point Chrges 25.4 Obtining the Vlue of the Electric Field from the Electric Potentil 25.5 Electric Potentil Due to Continuous Chrge Distributions 25.6 Electric Potentil Due to Chrged Conductor 25.7 (Optionl) The Millikn Oil-Drop Experiment 25.8 (Optionl) Applictions of Electrosttics 768

64 25.1 Potentil Difference nd Electric Potentil 769 T he concept of potentil energy ws introduced in Chpter 8 in connection with such conservtive forces s the force of grvity nd the elstic force exerted by spring. By using the lw of conservtion of energy, we were ble to void working directly with forces when solving vrious problems in mechnics. In this chpter we see tht the concept of potentil energy is lso of gret vlue in the study of electricity. Becuse the electrosttic force given by Coulomb s lw is conservtive, electrosttic phenomen cn be conveniently described in terms of n electric potentil energy. This ide enbles us to define sclr quntity known s electric potentil. Becuse the electric potentil t ny point in n electric field is sclr function, we cn use it to describe electrosttic phenomen more simply thn if we were to rely only on the concepts of the electric field nd electric forces. In lter chpters we shll see tht the concept of electric potentil is of gret prcticl vlue POTENTIAL DIFFERENCE AND ELECTRIC POTENTIAL When test chrge q 0 is plced in n electric field E creted by some other chrged object, the electric force cting on the test chrge is q 0 E. (If the field is produced by more thn one chrged object, this force cting on the test chrge is the vector sum of the individul forces exerted on it by the vrious other chrged objects.) The force q 0 E is conservtive becuse the individul forces described by Coulomb s lw re conservtive. When the test chrge is moved in the field by some externl gent, the work done by the field on the chrge is equl to the negtive of the work done by the externl gent cusing the displcement. For n infinitesiml displcement ds, the work done by the electric field on the chrge is F ds q 0 E ds. As this mount of work is done by the field, the potentil energy of the chrgefield system is decresed by n mount du q 0 E ds. For finite displcement of the chrge from point A to point B, the chnge in potentil energy of the system U U B U A is U q 0 B E ds (25.1) The integrtion is performed long the pth tht q 0 follows s it moves from A to B, nd the integrl is clled either pth integrl or line integrl (the two terms re synonymous). Becuse the force q 0 E is conservtive, this line integrl does not depend on the pth tken from A to B. A Chnge in potentil energy Quick Quiz 25.1 If the pth between A nd B does not mke ny difference in Eqution 25.1, why don t we just use the expression U q 0 Ed, where d is the stright-line distnce between A nd B? The potentil energy per unit chrge U/q 0 is independent of the vlue of q 0 nd hs unique vlue t every point in n electric field. This quntity U/q 0 is clled the electric potentil (or simply the potentil) V. Thus, the electric potentil t ny point in n electric field is V U q 0 (25.2)

65 770 CHAPTER 25 Electric Potentil The fct tht potentil energy is sclr quntity mens tht electric potentil lso is sclr quntity. The potentil difference V V B V A between ny two points A nd B in n electric field is defined s the chnge in potentil energy of the system divided by the test chrge q 0 : Potentil difference V U q 0 B E ds A (25.3) Potentil difference should not be confused with difference in potentil energy. The potentil difference is proportionl to the chnge in potentil energy, nd we see from Eqution 25.3 tht the two re relted by U q 0 V. Electric potentil is sclr chrcteristic of n electric field, independent of the chrges tht my be plced in the field. However, when we spek of potentil energy, we re referring to the chrgefield system. Becuse we re usully interested in knowing the electric potentil t the loction of chrge nd the potentil energy resulting from the interction of the chrge with the field, we follow the common convention of speking of the potentil energy s if it belonged to the chrge. Becuse the chnge in potentil energy of chrge is the negtive of the work done by the electric field on the chrge (s noted in Eqution 25.1), the potentil difference V between points A nd B equls the work per unit chrge tht n externl gent must perform to move test chrge from A to B without chnging the kinetic energy of the test chrge. Just s with potentil energy, only differences in electric potentil re meningful. To void hving to work with potentil differences, however, we often tke the vlue of the electric potentil to be zero t some convenient point in n electric field. This is wht we do here: rbitrrily estblish the electric potentil to be zero t point tht is infinitely remote from the chrges producing the field. Hving mde this choice, we cn stte tht the electric potentil t n rbitrry point in n electric field equls the work required per unit chrge to bring positive test chrge from infinity to tht point. Thus, if we tke point A in Eqution 25.3 to be t infinity, the electric potentil t ny point P is V P P E ds (25.4) In relity, V P represents the potentil difference V between the point P nd point t infinity. (Eq is specil cse of Eq ) Becuse electric potentil is mesure of potentil energy per unit chrge, the SI unit of both electric potentil nd potentil difference is joules per coulomb, which is defined s volt (V): Definition of volt 1 V 1 J C Tht is, 1 J of work must be done to move 1-C chrge through potentil difference of 1 V. Eqution 25.3 shows tht potentil difference lso hs units of electric field times distnce. From this, it follows tht the SI unit of electric field (N/C) cn lso be expressed in volts per meter: 1 N C 1 V m

66 25.2 Potentil Differences in Uniform Electric Field 771 A unit of energy commonly used in tomic nd nucler physics is the electron volt (ev), which is defined s the energy n electron (or proton) gins or loses by moving through potentil difference of 1 V. Becuse 1 V 1 J/C nd becuse the fundmentl chrge is pproximtely C, the electron volt is relted to the joule s follows: 1 ev CV J (25.5) For instnce, n electron in the bem of typicl television picture tube my hve speed of m/s. This corresponds to kinetic energy of J, which is equivlent to ev. Such n electron hs to be ccelerted from rest through potentil difference of 3.5 kv to rech this speed. The electron volt 25.2 POTENTIAL DIFFERENCES IN A UNIFORM ELECTRIC FIELD Equtions 25.1 nd 25.3 hold in ll electric fields, whether uniform or vrying, but they cn be simplified for uniform field. First, consider uniform electric field directed long the negtive y xis, s shown in Figure Let us clculte the potentil difference between two points A nd B seprted by distnce d, where d is mesured prllel to the field lines. Eqution 25.3 gives V B V A V B E ds B E cos 0 ds B E ds A Becuse E is constnt, we cn remove it from the integrl sign; this gives V E B ds Ed A (25.6) The minus sign indictes tht point B is t lower electric potentil thn point A; tht is, V B V A. Electric field lines lwys point in the direction of decresing electric potentil, s shown in Figure Now suppose tht test chrge q 0 moves from A to B. We cn clculte the chnge in its potentil energy from Equtions 25.3 nd 25.6: U q 0 V q 0 Ed (25.7) A A Potentil difference in uniform electric field A A E B q () d g B m (b) d Figure 25.1 () When the electric field E is directed downwrd, point B is t lower electric potentil thn point A. A positive test chrge tht moves from point A to point B loses electric potentil energy. (b) A mss m moving downwrd in the direction of the grvittionl field g loses grvittionl potentil energy.

67 772 CHAPTER 25 Electric Potentil QuickLb It tkes n electric field of bout V/cm to cuse sprk in dry ir. Shuffle cross rug nd rech towrd doorknob. By estimting the length of the sprk, determine the electric potentil difference between your finger nd the doorknob fter shuffling your feet but before touching the knob. (If it is very humid on the dy you ttempt this, it my not work. Why?) A Figure 25.2 A uniform electric field directed long the positive x xis. Point B is t lower electric potentil thn point A. Points B nd C re t the sme electric potentil. An equipotentil surfce s B C E 11.9 From this result, we see tht if q 0 is positive, then U is negtive. We conclude tht positive chrge loses electric potentil energy when it moves in the direction of the electric field. This mens tht n electric field does work on positive chrge when the chrge moves in the direction of the electric field. (This is nlogous to the work done by the grvittionl field on flling mss, s shown in Figure 25.1b.) If positive test chrge is relesed from rest in this electric field, it experiences n electric force q 0 E in the direction of E (downwrd in Fig. 25.1). Therefore, it ccelertes downwrd, gining kinetic energy. As the chrged prticle gins kinetic energy, it loses n equl mount of potentil energy. If q 0 is negtive, then U is positive nd the sitution is reversed: A negtive chrge gins electric potentil energy when it moves in the direction of the electric field. If negtive chrge is relesed from rest in the field E, it ccelertes in direction opposite the direction of the field. Now consider the more generl cse of chrged prticle tht is free to move between ny two points in uniform electric field directed long the x xis, s shown in Figure (In this sitution, the chrge is not being moved by n externl gent s before.) If s represents the displcement vector between points A nd B, Eqution 25.3 gives V B E ds E B ds Es A (25.8) where gin we re ble to remove E from the integrl becuse it is constnt. The chnge in potentil energy of the chrge is U q 0 V q 0 E s (25.9) Finlly, we conclude from Eqution 25.8 tht ll points in plne perpendiculr to uniform electric field re t the sme electric potentil. We cn see this in Figure 25.2, where the potentil difference V B V A is equl to the potentil difference V C V A. (Prove this to yourself by working out the dot product E s for s A:B, where the ngle between E nd s is rbitrry s shown in Figure 25.2, nd the dot product for s A:C, where 0.) Therefore, V B V C. The nme equipotentil surfce is given to ny surfce consisting of continuous distribution of points hving the sme electric potentil. Note tht becuse U q 0 V, no work is done in moving test chrge between ny two points on n equipotentil surfce. The equipotentil surfces of uniform electric field consist of fmily of plnes tht re ll perpendiculr to the field. Equipotentil surfces for fields with other symmetries re described in lter sections. A Quick Quiz 25.2 The lbeled points in Figure 25.3 re on series of equipotentil surfces ssocited with n electric field. Rnk (from gretest to lest) the work done by the electric field on positively chrged prticle tht moves from A to B; from B to C; from C to D; from D to E. 9 V 8 V 7 V A C B E D 6 V Figure 25.3 Four equipotentil surfces.

68 25.2 Potentil Differences in Uniform Electric Field 773 EXAMPLE 25.1 The Electric Field Between Two Prllel Pltes of Opposite Chrge A bttery produces specified potentil difference between conductors ttched to the bttery terminls. A 12-V bttery is connected between two prllel pltes, s shown in Figure The seprtion between the pltes is d 0.30 cm, nd we ssume the electric field between the pltes to be uniform. Figure 25.4 B d A 12 V A 12-V bttery connected to two prllel pltes. The electric field between the pltes hs mgnitude given by the potentil difference V divided by the plte seprtion d. (This ssumption is resonble if the plte seprtion is smll reltive to the plte dimensions nd if we do not consider points ner the plte edges.) Find the mgnitude of the electric field between the pltes. Solution The electric field is directed from the positive plte (A) to the negtive one (B ), nd the positive plte is t higher electric potentil thn the negtive plte is. The potentil difference between the pltes must equl the potentil difference between the bttery terminls. We cn understnd this by noting tht ll points on conductor in equilibrium re t the sme electric potentil 1 ; no potentil difference exists between terminl nd ny portion of the plte to which it is connected. Therefore, the mgnitude of the electric field between the pltes is, from Eqution 25.6, E V B V A d 12 V m V/m This configurtion, which is shown in Figure 25.4 nd clled prllel-plte cpcitor, is exmined in greter detil in Chpter 26. EXAMPLE 25.2 Motion of Proton in Uniform Electric Field A proton is relesed from rest in uniform electric field tht hs mgnitude of V/m nd is directed long the positive x xis (Fig. 25.5). The proton undergoes displcement of 0.50 m in the direction of E. () Find the chnge in electric potentil between points A nd B. Solution Becuse the proton (which, s you remember, crries positive chrge) moves in the direction of the field, we expect it to move to position of lower electric potentil. E From Eqution 25.6, we hve (b) Find the chnge in potentil energy of the proton for this displcement. Solution V Ed ( V/m)(0.50 m) U q 0 V e V ( C)( V) V J Figure 25.5 v va = 0 B B A d A proton ccelertes from A to B in the direction of the electric field. The negtive sign mens the potentil energy of the proton decreses s it moves in the direction of the electric field. As the proton ccelertes in the direction of the field, it gins kinetic energy nd t the sme time loses electric potentil energy (becuse energy is conserved). Exercise Use the concept of conservtion of energy to find the speed of the proton t point B. Answer m/s. 1 The electric field vnishes within conductor in electrosttic equilibrium; thus, the pth integrl E ds between ny two points in the conductor must be zero. A more complete discussion of this point is given in Section 25.6.

69 774 CHAPTER 25 Electric Potentil 25.3 ELECTRIC POTENTIAL AND POTENTIAL ENERGY DUE TO POINT CHARGES Consider n isolted positive point chrge q. Recll tht such chrge produces n electric field tht is directed rdilly outwrd from the chrge. To find the electric potentil t point locted distnce r from the chrge, we begin with the generl expression for potentil difference: V B V A B E ds A dr θ ds B where A nd B re the two rbitrry points shown in Figure At ny field point, the electric field due to the point chrge is E k e q rˆ/r 2 (Eq. 23.4), where rˆ is unit vector directed from the chrge towrd the field point. The quntity E ds cn be expressed s r r B E ds k q e r 2 rˆ ds A r A ˆr q Becuse the mgnitude of rˆ is 1, the dot product rˆ ds ds cos, where is the ngle between rˆ nd ds. Furthermore, ds cos is the projection of ds onto r; thus, ds cos dr. Tht is, ny displcement ds long the pth from point A to point B produces chnge dr in the mgnitude of r, the rdil distnce to the chrge creting the field. Mking these substitutions, we find tht E ds (k e q/r 2 )dr; hence, the expression for the potentil difference becomes Figure 25.6 The potentil difference between points A nd B due to point chrge q depends only on the initil nd finl rdil coordintes r A nd r B. The two dshed circles represent cross-sections of sphericl equipotentil surfces. Electric potentil creted by point chrge V B V A E r dr k e q r B V B V A k e q 1 1 (25.10) r B r A The integrl of E ds is independent of the pth between points A nd B s it must be becuse the electric field of point chrge is conservtive. Furthermore, Eqution expresses the importnt result tht the potentil difference between ny two points A nd B in field creted by point chrge depends only on the rdil coordintes r A nd r B. It is customry to choose the reference of electric potentil to be zero t r A. With this reference, the electric potentil creted by point chrge t ny distnce r from the chrge is V k e q r (25.11) Electric potentil is grphed in Figure 25.7 s function of r, the rdil distnce from positive chrge in the xy plne. Consider the following nlogy to grvittionl potentil: Imgine trying to roll mrble towrd the top of hill shped like Figure The grvittionl force experienced by the mrble is nlogous to the repulsive force experienced by positively chrged object s it pproches nother positively chrged object. Similrly, the electric potentil grph of the region surrounding negtive chrge is nlogous to hole with respect to ny pproching positively chrged objects. A chrged object must be infinitely distnt from nother chrge before the surfce is flt nd hs n electric potentil of zero. r A dr r 2 k eq r rb r A

70 25.3 Electric Potentil nd Potentil Energy Due to Point Chrges Electric potentil (V) y 0 x () Figure 25.7 () The electric potentil in the plne round single positive chrge is plotted on the verticl xis. (The electric potentil function for negtive chrge would look like hole insted of hill.) The red line shows the 1/r nture of the electric potentil, s given by Eqution (b) View looking stright down the verticl xis of the grph in prt (), showing concentric circles where the electric potentil is constnt. These circles re cross sections of equipotentil spheres hving the chrge t the center. (b)

71 776 CHAPTER 25 Electric Potentil Quick Quiz 25.3 A sphericl blloon contins positively chrged object t its center. As the blloon is inflted to greter volume while the chrged object remins t the center, does the electric potentil t the surfce of the blloon increse, decrese, or remin the sme? How bout the mgnitude of the electric field? The electric flux? Electric potentil due to severl point chrges Electric potentil energy due to two chrges We obtin the electric potentil resulting from two or more point chrges by pplying the superposition principle. Tht is, the totl electric potentil t some point P due to severl point chrges is the sum of the potentils due to the individul chrges. For group of point chrges, we cn write the totl electric potentil t P in the form q V k e i (25.12) i r i where the potentil is gin tken to be zero t infinity nd r i is the distnce from the point P to the chrge q i. Note tht the sum in Eqution is n lgebric sum of sclrs rther thn vector sum (which we use to clculte the electric field of group of chrges). Thus, it is often much esier to evlute V thn to evlute E. The electric potentil round dipole is illustrted in Figure We now consider the potentil energy of system of two chrged prticles. If V 1 is the electric potentil t point P due to chrge q 1, then the work n externl gent must do to bring second chrge q 2 from infinity to P without ccelertion is q 2 V 1. By definition, this work equls the potentil energy U of the two-prticle system when the prticles re seprted by distnce r 12 (Fig. 25.9). Therefore, we cn express the potentil energy s 2 U k q 1q 2 e (25.13) r 12 Note tht if the chrges re of the sme sign, U is positive. This is consistent with the fct tht positive work must be done by n externl gent on the system to bring the two chrges ner one nother (becuse like chrges repel). If the chrges re of opposite sign, U is negtive; this mens tht negtive work must be done ginst the ttrctive force between the unlike chrges for them to be brought ner ech other. If more thn two chrged prticles re in the system, we cn obtin the totl potentil energy by clculting U for every pir of chrges nd summing the terms lgebriclly. As n exmple, the totl potentil energy of the system of three chrges shown in Figure is U k e q 1 q 2 q 1q 3 q 2q 3 (25.14) r 12 r 13 r 23 Physiclly, we cn interpret this s follows: Imgine tht q 1 is fixed t the position shown in Figure but tht q 2 nd q 3 re t infinity. The work n externl gent must do to bring q 2 from infinity to its position ner q 1 is k e q 1 q 2 /r 12, which is the first term in Eqution The lst two terms represent the work required to bring q 3 from infinity to its position ner q 1 nd q 2. (The result is independent of the order in which the chrges re trnsported.) 2 The expression for the electric potentil energy of system mde up of two point chrges, Eqution 25.13, is of the sme form s the eqution for the grvittionl potentil energy of system mde up of two point msses, Gm 1 m 2 /r (see Chpter 14). The similrity is not surprising in view of the fct tht both expressions re derived from n inverse-squre force lw.

72 25.3 Electric Potentil nd Potentil Energy Due to Point Chrges r12 q 2 Electric potentil (V) q 1 Figure If two point chrges re seprted by distnce r 12, the potentil energy of the pir of chrges is given by k e q 1 q 2 /r 12. q 2 r 12 r y q 1 r 13 q x () Figure Three point chrges re fixed t the positions shown. The potentil energy of this system of chrges is given by Eqution Figure 25.8 () The electric potentil in the plne contining dipole. (b) Top view of the function grphed in prt (). (b)

73 778 CHAPTER 25 Electric Potentil EXAMPLE 25.3 The Electric Potentil Due to Two Point Chrges A chrge q C is locted t the origin, nd chrge q C is locted t (0, 3.00) m, s shown in Figure () Find the totl electric potentil due to these chrges t the point P, whose coordintes re (4.00, 0) m. Solution V P k e q 1 r 1 q 2 r 2 For two chrges, the sum in Eqution gives Nm2 C C m V C 5.00 m (b) Find the chnge in potentil energy of 3.00-C chrge s it moves from infinity to point P (Fig b). Solution When the chrge is t infinity, U i 0, nd when the chrge is t P, U f q 3 V P ; therefore, Therefore, becuse W U, positive work would hve to be done by n externl gent to remove the chrge from point P bck to infinity. Exercise U q 3 V P 0 ( C)( V) Find the totl potentil energy of the system illustrted in Figure 25.11b. Answer J J y y 6.00 µc µ 6.00 µc µ 3.00 m 3.00 m P x 2.00 µc µ 4.00 m 2.00 µc µ 4.00 m 3.00 µc µ x () (b) Figure () The electric potentil t P due to the two chrges is the lgebric sum of the potentils due to the individul chrges. (b) Wht is the potentil energy of the three-chrge system? 25.4 OBTAINING THE VALUE OF THE ELECTRIC FIELD FROM THE ELECTRIC POTENTIAL The electric field E nd the electric potentil V re relted s shown in Eqution We now show how to clculte the vlue of the electric field if the electric potentil is known in certin region. From Eqution 25.3 we cn express the potentil difference dv between two points distnce ds prt s dv Eds (25.15) If the electric field hs only one component E x, then E ds E x dx. Therefore, Eqution becomes dv E x dx, or E x dv dx (25.16)

74 25.4 Obtining the Vlue of the Electric Field from the Electric Potentil 779 Tht is, the mgnitude of the electric field in the direction of some coordinte is equl to the negtive of the derivtive of the electric potentil with respect to tht coordinte. Recll from the discussion following Eqution 25.8 tht the electric potentil does not chnge for ny displcement perpendiculr to n electric field. This is consistent with the notion, developed in Section 25.2, tht equipotentil surfces re perpendiculr to the field, s shown in Figure A smll positive chrge plced t rest on n electric field line begins to move long the direction of E becuse tht is the direction of the force exerted on the chrge by the chrge distribution creting the electric field (nd hence is the direction of ). Becuse the chrge strts with zero velocity, it moves in the direction of the chnge in velocity tht is, in the direction of. In Figures nd 25.12b, chrge plced t rest in the field will move in stright line becuse its ccelertion vector is lwys prllel to its velocity vector. The mgnitude of v increses, but its direction does not chnge. The sitution is different in Figure 25.12c. A positive chrge plced t some point ner the dipole first moves in direction prllel to E t tht point. Becuse the direction of the electric field is different t different loctions, however, the force cting on the chrge chnges direction, nd is no longer prllel to v. This cuses the moving chrge to chnge direction nd speed, but it does not necessrily follow the electric field lines. Recll tht it is not the velocity vector but rther the ccelertion vector tht is proportionl to force. If the chrge distribution creting n electric field hs sphericl symmetry such tht the volume chrge density depends only on the rdil distnce r, then the electric field is rdil. In this cse, E ds E r dr, nd thus we cn express dv in the form dv E r dr. Therefore, E r dv (25.17) dr For exmple, the electric potentil of point chrge is V k e q/r. Becuse V is function of r only, the potentil function hs sphericl symmetry. Applying Eqution 25.17, we find tht the electric field due to the point chrge is E r k e q/r 2, fmilir result. Note tht the potentil chnges only in the rdil direction, not in q E () (b) Figure Equipotentil surfces (dshed blue lines) nd electric field lines (red lines) for () uniform electric field produced by n infinite sheet of chrge, (b) point chrge, nd (c) n electric dipole. In ll cses, the equipotentil surfces re perpendiculr to the electric field lines t every point. Compre these drwings with Figures 25.2, 25.7b, nd 25.8b. (c)

75 780 CHAPTER 25 Electric Potentil Equipotentil surfces re perpendiculr to the electric field lines ny direction perpendiculr to r. Thus, V (like E r ) is function only of r. Agin, this is consistent with the ide tht equipotentil surfces re perpendiculr to field lines. In this cse the equipotentil surfces re fmily of spheres concentric with the sphericlly symmetric chrge distribution (Fig b). The equipotentil surfces for n electric dipole re sketched in Figure 25.12c. When test chrge undergoes displcement ds long n equipotentil surfce, then dv 0 becuse the potentil is constnt long n equipotentil surfce. From Eqution 25.15, then, dv Eds 0; thus, E must be perpendiculr to the displcement long the equipotentil surfce. This shows tht the equipotentil surfces must lwys be perpendiculr to the electric field lines. In generl, the electric potentil is function of ll three sptil coordintes. If V(r) is given in terms of the crtesin coordintes, the electric field components E x, E y, nd E z cn redily be found from V(x, y, z) s the prtil derivtives 3 E x V x E y V y E z V z For exmple, if V 3x 2 y y 2 yz, then V x x (3x2 y y 2 yz) x (3x2 y) 3y d dx (x2 ) 6xy EXAMPLE 25.4 The Electric Potentil Due to Dipole An electric dipole consists of two chrges of equl mgnitude nd opposite sign seprted by distnce 2, s shown in Figure The dipole is long the x xis nd is centered t the origin. () Clculte the electric potentil t point P. Solution For point P in Figure 25.13, (How would this result chnge if point P hppened to be locted to the left of the negtive chrge?) (b) Clculte V nd E x t point fr from the dipole. Solution If point P is fr from the dipole, such tht x W, then 2 cn be neglected in the term x 2 2, nd V becomes V k e q i r i k e q x q x 2k e q x 2 2 V 2k e q x 2 (x W ) Using Eqution nd this result, we cn clculte the electric field t point fr from the dipole: y E x dv dx 4k e q x 3 ( x W ) Figure P x q q x An electric dipole locted on the x xis. (c) Clculte V nd E x if point P is locted nywhere between the two chrges. Solution V k e q i r i k e q x E x dv dx d dx q x 2k eqx x 2 2 2k eqx 2 x 2 2k eq x (x 2 2 ) 3 In vector nottion, E is often written where is clled the grdient opertor. E V i x j y k z V

76 25.5 Electric Potentil Due to Continuous Chrge Distributions 781 We cn check these results by considering the sitution t the center of the dipole, where x 0, V 0, nd E x 2k e q/ 2. Exercise Verify the electric field result in prt (c) by clculting the sum of the individul electric field vectors t the origin due to the two chrges ELECTRIC POTENTIAL DUE TO CONTINUOUS CHARGE DISTRIBUTIONS We cn clculte the electric potentil due to continuous chrge distribution in two wys. If the chrge distribution is known, we cn strt with Eqution for the electric potentil of point chrge. We then consider the potentil due to smll chrge element dq, treting this element s point chrge (Fig ). The electric potentil dv t some point P due to the chrge element dq is dv k dq e (25.18) r where r is the distnce from the chrge element to point P. To obtin the totl potentil t point P, we integrte Eqution to include contributions from ll elements of the chrge distribution. Becuse ech element is, in generl, different distnce from point P nd becuse k e is constnt, we cn express V s V k e dq (25.19) r In effect, we hve replced the sum in Eqution with n integrl. Note tht this expression for V uses prticulr reference: The electric potentil is tken to be zero when point P is infinitely fr from the chrge distribution. If the electric field is lredy known from other considertions, such s Guss s lw, we cn clculte the electric potentil due to continuous chrge distribution using Eqution If the chrge distribution is highly symmetric, we first evlute E t ny point using Guss s lw nd then substitute the vlue obtined into Eqution 25.3 to determine the potentil difference V between ny two points. We then choose the electric potentil V to be zero t some convenient point. We illustrte both methods with severl exmples. P r Figure dq The electric potentil t the point P due to continuous chrge distribution cn be clculted by dividing the chrged body into segments of chrge dq nd summing the electric potentil contributions over ll segments. EXAMPLE 25.5 Electric Potentil Due to Uniformly Chrged Ring () Find n expression for the electric potentil t point P locted on the perpendiculr centrl xis of uniformly chrged ring of rdius nd totl chrge Q. Solution Let us orient the ring so tht its plne is perpendiculr to n x xis nd its center is t the origin. We cn then tke point P to be t distnce x from the center of the ring, s shown in Figure The chrge element dq is t distnce!x 2 2 from point P. Hence, we cn express V s V k e dq dq k r e!x 2 2 Becuse ech element dq is t the sme distnce from point P, we cn remove!x 2 2 from the integrl, nd V reduces to (25.20) The only vrible in this expression for V is x. This is not surprising becuse our clcultion is vlid only for points long the x xis, where y nd z re both zero. (b) Find n expression for the mgnitude of the electric field t point P. Solution V k e!x 2 2 dq From symmetry, we see tht long the x xis E cn hve only n x component. Therefore, we cn use Equk e Q!x 2 2

77 782 CHAPTER 25 Electric Potentil tion 25.16: E x dv dx k eq d dx (x2 2 ) 1/2 k e Q( 1 2 )(x2 2 ) 3/2 (2x) ter, V hs either mximum or minimum vlue; it is, in fct, mximum. dq k e Qx (x 2 2 ) 3/2 (25.21) x 2 2 This result grees with tht obtined by direct integrtion (see Exmple 23.8). Note tht E x 0 t x 0 (the center of the ring). Could you hve guessed this from Coulomb s lw? Exercise Wht is the electric potentil t the center of the ring? Wht does the vlue of the field t the center tell you bout the vlue of V t the center? Answer V k e Q /. Becuse E x dv/dx 0 t the cen- Figure A uniformly chrged ring of rdius lies in plne perpendiculr to the x xis. All segments dq of the ring re the sme distnce from ny point P lying on the x xis. x P EXAMPLE 25.6 Electric Potentil Due to Uniformly Chrged Disk Find () the electric potentil nd (b) the mgnitude of the electric field long the perpendiculr centrl xis of uniformly chrged disk of rdius nd surfce chrge density. Solution () Agin, we choose the point P to be t distnce x from the center of the disk nd tke the plne of the disk to be perpendiculr to the x xis. We cn simplify the problem by dividing the disk into series of chrged rings. The electric potentil of ech ring is given by Eqution Consider one such ring of rdius r nd width dr, s indicted in Figure The surfce re of the ring is da 2r dr; Figure dr r r 2 x 2 A uniformly chrged disk of rdius lies in plne perpendiculr to the x xis. The clcultion of the electric potentil t ny point P on the x xis is simplified by dividing the disk into mny rings ech of re 2r dr. x da = 2πrdrπ P from the definition of surfce chrge density (see Section 23.5), we know tht the chrge on the ring is dq Hence, the potentil t the point P due to this ring is da 2r dr. To find the totl electric potentil t P, we sum over ll rings mking up the disk. Tht is, we integrte dv from r 0 to r : V k e 0 dv This integrl is of the form u n du nd hs the vlue u n1 /(n 1), where n 1 2 nd u r 2 x 2. This gives V (25.22) (b) As in Exmple 25.5, we cn find the electric field t ny xil point from E x dv dx k e dq!r 2 x k e 2r dr 2!r 2 x 2 2r dr!r 2 x k e (r 2 x 2 ) 1/2 2r dr 2 0 2k e [(x 2 2 ) 1/2 x] 2k e 1 x 2!x 2 (25.23) The clcultion of V nd E for n rbitrry point off the xis is more difficult to perform, nd we do not tret this sitution in this text.

78 25.5 Electric Potentil Due to Continuous Chrge Distributions 783 EXAMPLE 25.7 Electric Potentil Due to Finite Line of Chrge A rod of length locted long the x xis hs totl chrge Q nd uniform liner chrge density Q /. Find the electric potentil t point P locted on the y xis distnce from the origin (Fig ). Evluting V, we find tht V k e Q ln!2 2 (25.24) Solution The length element dx hs chrge dq dx. Becuse this element is distnce r!x 2 2 from point P, we cn express the potentil t point P due to this element s To obtin the totl potentil t P, we integrte this expression over the limits x 0 to x. Noting tht k e nd re constnts, we find tht V k e This integrl hs the following vlue (see Appendix B): dv k e dq r 0 dx!x 2 k Q 2 e 0 dx k e dx!x 2 2!x 2 2 ln(x!x2 2 ) dx!x 2 2 P 0 y x r Figure A uniform line chrge of length locted long the x xis. To clculte the electric potentil t P, the line chrge is divided into segments ech of length dx nd ech crrying chrge dq dx. dq dx x EXAMPLE 25.8 Electric Potentil Due to Uniformly Chrged Sphere An insulting solid sphere of rdius R hs uniform positive volume chrge density nd totl chrge Q. () Find the electric potentil t point outside the sphere, tht is, for r R. Tke the potentil to be zero t r. Solution In Exmple 24.5, we found tht the mgnitude of the electric field outside uniformly chrged sphere of rdius R is E r k e Q r 2 (for r R ) Becuse the potentil must be continuous t r R, we cn use this expression to obtin the potentil t the surfce of the sphere. Tht is, the potentil t point such s C shown in Figure is V C k e Q R (for r R ) (b) Find the potentil t point inside the sphere, tht is, for r R. where the field is directed rdilly outwrd when Q is positive. In this cse, to obtin the electric potentil t n exterior point, such s B in Figure 25.18, we use Eqution 25.4 nd the expression for E r given bove: V B r E r dr k e Q r dr r 2 Q R r D C B V B k e Q r (for r R ) Note tht the result is identicl to the expression for the electric potentil due to point chrge (Eq ). Figure A uniformly chrged insulting sphere of rdius R nd totl chrge Q. The electric potentils t points B nd C re equivlent to those produced by point chrge Q locted t the center of the sphere, but this is not true for point D.

79 784 CHAPTER 25 Electric Potentil Solution In Exmple 24.5 we found tht the electric field inside n insulting uniformly chrged sphere is (for r R ) We cn use this result nd Eqution 25.3 to evlute the potentil difference V D V C t some interior point D: V D V C r E r dr k eq Substituting V C k e Q /R into this expression nd solving for V D, we obtin (for r R ) (25.25) At r R, this expression gives result tht grees with tht for the potentil t the surfce, tht is, V C. A plot of V versus r for this chrge distribution is given in Figure Exercise E r k eq R 3 R R 3 r R V D k eq 2R 3 r 2 R 2 r r dr k eq 2R 3 (R 2 r 2 ) Wht re the mgnitude of the electric field nd the electric potentil t the center of the sphere? Answer E 0; V 0 3k e Q /2R. Figure V 0 V 0 V V 0 = 3k e Q 2R V D = k e Q 2R (3 r 2 R 2 ) R V B = k eq r A plot of electric potentil V versus distnce r from the center of uniformly chrged insulting sphere of rdius R. The curve for V D inside the sphere is prbolic nd joins smoothly with the curve for V B outside the sphere, which is hyperbol. The potentil hs mximum vlue V 0 t the center of the sphere. We could mke this grph three dimensionl (similr to Figures 25.7 nd 25.8) by spinning it round the verticl xis. r 25.6 ELECTRIC POTENTIAL DUE TO A CHARGED CONDUCTOR In Section 24.4 we found tht when solid conductor in equilibrium crries net chrge, the chrge resides on the outer surfce of the conductor. Furthermore, we showed tht the electric field just outside the conductor is perpendiculr to the surfce nd tht the field inside is zero. We now show tht every point on the surfce of chrged conductor in equilibrium is t the sme electric potentil. Consider two points A nd B on the surfce of chrged conductor, s shown in Figure Along surfce pth connecting these points, E is lwys perpendiculr to the displcement ds; there- B A E Figure An rbitrrily shped conductor crrying positive chrge. When the conductor is in electrosttic equilibrium, ll of the chrge resides t the surfce, E 0 inside the conductor, nd the direction of E just outside the conductor is perpendiculr to the surfce. The electric potentil is constnt inside the conductor nd is equl to the potentil t the surfce. Note from the spcing of the plus signs tht the surfce chrge density is nonuniform.

80 25.6 Electric Potentil Due to Chrged Conductor 785 fore E ds 0. Using this result nd Eqution 25.3, we conclude tht the potentil difference between A nd B is necessrily zero: V B V A B E ds 0 This result pplies to ny two points on the surfce. Therefore, V is constnt everywhere on the surfce of chrged conductor in equilibrium. Tht is, A the surfce of ny chrged conductor in electrosttic equilibrium is n equipotentil surfce. Furthermore, becuse the electric field is zero inside the conductor, we conclude from the reltionship E r dv/dr tht the electric potentil is constnt everywhere inside the conductor nd equl to its vlue t the surfce. The surfce of chrged conductor is n equipotentil surfce Becuse this is true bout the electric potentil, no work is required to move test chrge from the interior of chrged conductor to its surfce. Consider solid metl conducting sphere of rdius R nd totl positive chrge Q, s shown in Figure The electric field outside the sphere is k e Q /r 2 nd points rdilly outwrd. From Exmple 25.8, we know tht the electric potentil t the interior nd surfce of the sphere must be k e Q /R reltive to infinity. The potentil outside the sphere is k e Q /r. Figure 25.21b is plot of the electric potentil s function of r, nd Figure 25.21c shows how the electric field vries with r. When net chrge is plced on sphericl conductor, the surfce chrge density is uniform, s indicted in Figure However, if the conductor is nonsphericl, s in Figure 25.20, the surfce chrge density is high where the rdius of curvture is smll nd the surfce is convex (s noted in Section 24.4), nd it is low where the rdius of curvture is smll nd the surfce is concve. Becuse the electric field just outside the conductor is proportionl to the surfce chrge density, we see tht the electric field is lrge ner convex points hving smll rdii of curvture nd reches very high vlues t shrp points. Figure shows the electric field lines round two sphericl conductors: one crrying net chrge Q, nd lrger one crrying zero net chrge. In this cse, the surfce chrge density is not uniform on either conductor. The sphere hving zero net chrge hs negtive chrges induced on its side tht fces the () R (b) (c) k e Q R V E k e Q r k e Q r 2 r R r Electric field pttern of chrged conducting plte plced ner n oppositely chrged pointed conductor. Smll pieces of thred suspended in oil lign with the electric field lines. The field surrounding the pointed conductor is most intense ner the pointed end nd t other plces where the rdius of curvture is smll. Figure () The excess chrge on conducting sphere of rdius R is uniformly distributed on its surfce. (b) Electric potentil versus distnce r from the center of the chrged conducting sphere. (c) Electric field mgnitude versus distnce r from the center of the chrged conducting sphere.

81 786 CHAPTER 25 Electric Potentil Q Q = 0 Figure The electric field lines (in red) round two sphericl conductors. The smller sphere hs net chrge Q, nd the lrger one hs zero net chrge. The blue curves re crosssections of equipotentil surfces. chrged sphere nd positive chrges induced on its side opposite the chrged sphere. The blue curves in the figure represent the cross-sections of the equipotentil surfces for this chrge configurtion. As usul, the field lines re perpendiculr to the conducting surfces t ll points, nd the equipotentil surfces re perpendiculr to the field lines everywhere. Trying to move positive chrge in the region of these conductors would be like moving mrble on hill tht is flt on top (representing the conductor on the left) nd hs nother flt re prtwy down the side of the hill (representing the conductor on the right). EXAMPLE 25.9 Two Connected Chrged Spheres Two sphericl conductors of rdii r 1 nd r 2 re seprted by distnce much greter thn the rdius of either sphere. The spheres re connected by conducting wire, s shown in Figure The chrges on the spheres in equilibrium re q 1 nd q 2, respectively, nd they re uniformly chrged. Find the rtio of the mgnitudes of the electric fields t the surfces of the spheres. q 1 r 1 Solution Becuse the spheres re connected by conducting wire, they must both be t the sme electric potentil: q 2 r 2 V k e q 1 r 1 k e q 2 r 2 Therefore, the rtio of chrges is Figure Two chrged sphericl conductors connected by conducting wire. The spheres re t the sme electric potentil V.

82 25.6 Electric Potentil Due to Chrged Conductor 787 (1) q 1 q 2 r 1 r 2 Tking the rtio of these two fields nd mking use of Eqution (1), we find tht Becuse the spheres re very fr prt nd their surfces uniformly chrged, we cn express the mgnitude of the electric fields t their surfces s E 1 k q 1 e r 2 1 nd E 2 k q 2 e r 2 2 E 1 E 2 r 2 r 1 Hence, the field is more intense in the vicinity of the smller sphere even though the electric potentils of both spheres re the sme. A Cvity Within Conductor Now consider conductor of rbitrry shpe contining cvity s shown in Figure Let us ssume tht no chrges re inside the cvity. In this cse, the electric field inside the cvity must be zero regrdless of the chrge distribution on the outside surfce of the conductor. Furthermore, the field in the cvity is zero even if n electric field exists outside the conductor. To prove this point, we use the fct tht every point on the conductor is t the sme electric potentil, nd therefore ny two points A nd B on the surfce of the cvity must be t the sme potentil. Now imgine tht field E exists in the cvity nd evlute the potentil difference V B V A defined by Eqution 25.3: V B V A B E ds If E is nonzero, we cn lwys find pth between A nd B for which E ds is positive number; thus, the integrl must be positive. However, becuse V B V A 0, the integrl of E ds must be zero for ll pths between ny two points on the conductor, which implies tht E is zero everywhere. This contrdiction cn be reconciled only if E is zero inside the cvity. Thus, we conclude tht cvity surrounded by conducting wlls is field-free region s long s no chrges re inside the cvity. A B Figure A conductor in electrosttic equilibrium contining cvity. The electric field in the cvity is zero, regrdless of the chrge on the conductor. A Coron Dischrge A phenomenon known s coron dischrge is often observed ner conductor such s high-voltge power line. When the electric field in the vicinity of the conductor is sufficiently strong, electrons re stripped from ir molecules. This cuses the molecules to be ionized, thereby incresing the ir s bility to conduct. The observed glow (or coron dischrge) results from the recombintion of free electrons with the ionized ir molecules. If conductor hs n irregulr shpe, the electric field cn be very high ner shrp points or edges of the conductor; consequently, the ioniztion process nd coron dischrge re most likely to occur round such points. Quick Quiz 25.4 () Is it possible for the mgnitude of the electric field to be zero t loction where the electric potentil is not zero? (b) Cn the electric potentil be zero where the electric field is nonzero?

83 788 CHAPTER 25 Electric Potentil v F D q mg Optionl Section 25.7 THE MILLIKAN OIL-DROP EXPERIMENT During the period from 1909 to 1913, Robert Millikn performed brillint set of experiments in which he mesured e, the elementry chrge on n electron, nd demonstrted the quntized nture of this chrge. His pprtus, digrmmed in Figure 25.25, contins two prllel metllic pltes. Chrged oil droplets from n tomizer re llowed to pss through smll hole in the upper plte. A horizontlly directed light bem (not shown in the digrm) is used to illuminte the oil droplets, which re viewed through telescope whose long xis is t right ngles to the light bem. When the droplets re viewed in this mnner, they pper s shining strs ginst drk bckground, nd the rte t which individul drops fll cn be determined. 4 Let us ssume tht single drop hving mss m nd crrying chrge q is being viewed nd tht its chrge is negtive. If no electric field is present between the pltes, the two forces cting on the chrge re the force of grvity mg cting downwrd nd viscous drg force F D cting upwrd s indicted in Figure The drg force is proportionl to the drop s speed. When the drop reches its terminl speed v, the two forces blnce ech other (mg F D ). Now suppose tht bttery connected to the pltes sets up n electric field between the pltes such tht the upper plte is t the higher electric potentil. In this cse, third force qe cts on the chrged drop. Becuse q is negtive nd E is directed downwrd, this electric force is directed upwrd, s shown in Figure 25.26b. If this force is sufficiently gret, the drop moves upwrd nd the drg force F D cts downwrd. When the upwrd electric force q E blnces the sum of the grvittionl force nd the downwrd drg force F D, the drop reches new terminl speed v in the upwrd direction. With the field turned on, drop moves slowly upwrd, typiclly t rtes of hundredths of centimeter per second. The rte of fll in the bsence of field is comprble. Hence, one cn follow single droplet for hours, lterntely rising nd flling, by simply turning the electric field on nd off. () Field off Oil droplets Atomizer qe Bttery Pin hole v E q Chrged plte v Chrged plte Telescope mg Figure (b) Field on F D The forces cting on negtively chrged oil droplet in the Millikn experiment. Figure Switch Schemtic drwing of the Millikn oil-drop pprtus. 4 At one time, the oil droplets were termed Millikn s Shining Strs. Perhps this description hs lost its populrity becuse of the genertions of physics students who hve experienced hllucintions, ner blindness, migrine hedches, nd so forth, while repeting Millikn s experiment!

84 25.8 Applictions of Electrosttics 789 After recording mesurements on thousnds of droplets, Millikn nd his coworkers found tht ll droplets, to within bout 1% precision, hd chrge equl to some integer multiple of the elementry chrge e : q ne n 0, 1, 2, 3,... where e C. Millikn s experiment yields conclusive evidence tht chrge is quntized. For this work, he ws wrded the Nobel Prize in Physics in Optionl Section 25.8 APPLICATIONS OF ELECTROSTATICS The prcticl ppliction of electrosttics is represented by such devices s lightning rods nd electrosttic precipittors nd by such processes s xerogrphy nd the pinting of utomobiles. Scientific devices bsed on the principles of electrosttics include electrosttic genertors, the field-ion microscope, nd ion-drive rocket engines. The Vn de Grff Genertor In Section 24.5 we described n experiment tht demonstrtes method for trnsferring chrge to hollow conductor (the Frdy ice-pil experiment). When chrged conductor is plced in contct with the inside of hollow conductor, ll of the chrge of the chrged conductor is trnsferred to the hollow conductor. In principle, the chrge on the hollow conductor nd its electric potentil cn be incresed without limit by repetition of the process. In 1929 Robert J. Vn de Grff ( ) used this principle to design nd build n electrosttic genertor. This type of genertor is used extensively in nucler physics reserch. A schemtic representtion of the genertor is given in Figure Chrge is delivered continuously to high-potentil electrode by mens of moving belt of insulting mteril. The high-voltge electrode is hollow conductor mounted on n insulting column. The belt is chrged t point A by mens of coron dischrge between comb-like metllic needles nd grounded grid. The needles re mintined t positive electric potentil of typiclly 10 4 V. The positive chrge on the moving belt is trnsferred to the hollow conductor by second comb of needles t point B. Becuse the electric field inside the hollow conductor is negligible, the positive chrge on the belt is esily trnsferred to the conductor regrdless of its potentil. In prctice, it is possible to increse the electric potentil of the hollow conductor until electricl dischrge occurs through the ir. Becuse the brekdown electric field in ir is bout V/m, sphere 1 m in rdius cn be rised to mximum potentil of V. The potentil cn be incresed further by incresing the rdius of the hollow conductor nd by plcing the entire system in continer filled with high-pressure gs. Vn de Grff genertors cn produce potentil differences s lrge s 20 million volts. Protons ccelerted through such lrge potentil differences receive enough energy to initite nucler rections between themselves nd vrious trget nuclei. Smller genertors re often seen in science clssrooms nd museums. If person insulted from the ground touches the sphere of Vn de Grff genertor, his or her body cn be brought to high electric potentil. The hir cquires net positive chrge, nd ech strnd is repelled by ll the others. The result is Grounded grid Figure B A Ground Hollow conductor Belt Schemtic digrm of Vn de Grff genertor. Chrge is trnsferred to the hollow conductor t the top by mens of moving belt. The chrge is deposited on the belt t point A nd trnsferred to the hollow conductor t point B. Insultor

85 790 CHAPTER 25 Electric Potentil scene such s tht depicted in the photogrph t the beginning of this chpter. In ddition to being insulted from ground, the person holding the sphere is sfe in this demonstrtion becuse the totl chrge on the sphere is very smll (on the order of 1 C). If this mount of chrge ccidentlly pssed from the sphere through the person to ground, the corresponding current would do no hrm. QuickLb Sprinkle some slt nd pepper on n open dish nd mix the two together. Now pull comb through your hir severl times nd bring the comb to within 1 cm of the slt nd pepper. Wht hppens? How is wht hppens here relted to the opertion of n electrosttic precipittor? The Electrosttic Precipittor One importnt ppliction of electricl dischrge in gses is the electrosttic precipittor. This device removes prticulte mtter from combustion gses, thereby reducing ir pollution. Precipittors re especilly useful in col-burning power plnts nd in industril opertions tht generte lrge quntities of smoke. Current systems re ble to eliminte more thn 99% of the sh from smoke. Figure shows schemtic digrm of n electrosttic precipittor. A high potentil difference (typiclly 40 to 100 kv) is mintined between wire running down the center of duct nd the wlls of the duct, which re grounded. The wire is mintined t negtive electric potentil with respect to the wlls, so the electric field is directed towrd the wire. The vlues of the field ner the wire become high enough to cuse coron dischrge round the wire; the dischrge ionizes some ir molecules to form positive ions, electrons, nd such negtive ions s O 2. The ir to be clened enters the duct nd moves ner the wire. As the electrons nd negtive ions creted by the dischrge re ccelerted towrd the outer wll by the electric field, the dirt prticles in the ir become chrged by collisions nd ion cpture. Becuse most of the chrged dirt prticles re negtive, they too re drwn to the duct wlls by the electric field. When the duct is periodiclly shken, the prticles brek loose nd re collected t the bottom. Insultor Clen ir out Dirty ir in Weight Dirt out () Figure (b) () Schemtic digrm of n electrosttic precipittor. The high negtive electric potentil mintined on the centrl coiled wire cretes n electricl dischrge in the vicinity of the wire. Compre the ir pollution when the electrosttic precipittor is (b) operting nd (c) turned off. (c)

86 25.8 Applictions of Electrosttics 791 In ddition to reducing the level of prticulte mtter in the tmosphere (compre Figs b nd c), the electrosttic precipittor recovers vluble mterils in the form of metl oxides. Xerogrphy nd Lser Printers The bsic ide of xerogrphy 5 ws developed by Chester Crlson, who ws grnted ptent for the xerogrphic process in The one feture of this process tht mkes it unique is the use of photoconductive mteril to form n imge. (A photoconductor is mteril tht is poor electricl conductor in the drk but tht becomes good electricl conductor when exposed to light.) The xerogrphic process is illustrted in Figure to d. First, the surfce of plte or drum tht hs been coted with thin film of photoconductive mteril (usully selenium or some compound of selenium) is given positive electrosttic chrge in the drk. An imge of the pge to be copied is then focused by lens onto the chrged surfce. The photoconducting surfce becomes conducting only in res where light strikes it. In these res, the light produces chrge crriers in the photoconductor tht move the positive chrge off the drum. However, positive Lens Light cuses some res of drum to become electriclly conducting, removing positive chrge Selenium-coted drum () Chrging the drum (b) Imging the document Negtively chrged toner (c) Applying the toner Interlced pttern of lser lines Lser bem Figure (d) Trnsferring the toner to the pper (e) Lser printer drum The xerogrphic process: () The photoconductive surfce of the drum is positively chrged. (b) Through the use of light source nd lens, n imge is formed on the surfce in the form of positive chrges. (c) The surfce contining the imge is covered with negtively chrged powder, which dheres only to the imge re. (d) A piece of pper is plced over the surfce nd given positive chrge. This trnsfers the imge to the pper s the negtively chrged powder prticles migrte to the pper. The pper is then het-treted to fix the powder. (e) A lser printer opertes similrly except the imge is produced by turning lser bem on nd off s it sweeps cross the selenium-coted drum. 5 The prefix xero- is from the Greek word mening dry. Note tht no liquid ink is used nywhere in xerogrphy.

87 792 CHAPTER 25 Electric Potentil chrges remin on those res of the photoconductor not exposed to light, leving ltent imge of the object in the form of positive surfce chrge distribution. Next, negtively chrged powder clled toner is dusted onto the photoconducting surfce. The chrged powder dheres only to those res of the surfce tht contin the positively chrged imge. At this point, the imge becomes visible. The toner (nd hence the imge) re then trnsferred to the surfce of sheet of positively chrged pper. Finlly, the toner is fixed to the surfce of the pper s the toner melts while pssing through high-temperture rollers. This results in permnent copy of the originl. A lser printer (Fig e) opertes by the sme principle, with the exception tht computer-directed lser bem is used to illuminte the photoconductor insted of lens. SUMMARY When positive test chrge q 0 is moved between points A nd B in n electric field E, the chnge in the potentil energy is U q 0 B E ds (25.1) The electric potentil V U/q 0 is sclr quntity nd hs units of joules per coulomb ( J/C), where 1 J/C 1 V. The potentil difference V between points A nd B in n electric field E is defined s V U (25.3) q 0 B E ds A The potentil difference between two points A nd B in uniform electric field E is V Ed (25.6) where d is the mgnitude of the displcement in the direction prllel to E. An equipotentil surfce is one on which ll points re t the sme electric potentil. Equipotentil surfces re perpendiculr to electric field lines. If we define V 0 t r A, the electric potentil due to point chrge t ny distnce r from the chrge is A V k e q r (25.11) We cn obtin the electric potentil ssocited with group of point chrges by summing the potentils due to the individul chrges. The potentil energy ssocited with pir of point chrges seprted by distnce r 12 is U k q 1q 2 e (25.13) r 12 This energy represents the work required to bring the chrges from n infinite seprtion to the seprtion r 12. We obtin the potentil energy of distribution of point chrges by summing terms like Eqution over ll pirs of prticles.

88 Summry 793 TABLE 25.1 Electric Potentil Due to Vrious Chrge Distributions Chrge Distribution Electric Potentil Loction Uniformly chrged ring of rdius Uniformly chrged disk of rdius Uniformly chrged, insulting solid sphere of rdius R nd totl chrge Q Isolted conducting sphere of rdius R nd totl chrge Q V k e V k Q e r V k Q e R Q!x 2 2 V 2k e [(x 2 2 ) 1/2 x] V k Q e r V k eq 2R 3 r 2 R 2 Along perpendiculr centrl xis of ring, distnce x from ring center Along perpendiculr centrl xis of disk, distnce x from disk center r R r R r R r R If we know the electric potentil s function of coordintes x, y, z, we cn obtin the components of the electric field by tking the negtive derivtive of the electric potentil with respect to the coordintes. For exmple, the x component of the electric field is E x dv (25.16) dx The electric potentil due to continuous chrge distribution is V k e dq (25.19) r Every point on the surfce of chrged conductor in electrosttic equilibrium is t the sme electric potentil. The potentil is constnt everywhere inside the conductor nd equl to its vlue t the surfce. Tble 25.1 lists electric potentils due to severl chrge distributions. Problem-Solving Hints Clculting Electric Potentil Remember tht electric potentil is sclr quntity, so components need not be considered. Therefore, when using the superposition principle to evlute the electric potentil t point due to system of point chrges, simply tke the lgebric sum of the potentils due to the vrious chrges. However, you must keep trck of signs. The potentil is positive for positive chrges, nd it is negtive for negtive chrges. Just s with grvittionl potentil energy in mechnics, only chnges in electric potentil re significnt; hence, the point where you choose the poten-

89 794 CHAPTER 25 Electric Potentil til to be zero is rbitrry. When deling with point chrges or chrge distribution of finite size, we usully define V 0 to be t point infinitely fr from the chrges. You cn evlute the electric potentil t some point P due to continuous distribution of chrge by dividing the chrge distribution into infinitesiml elements of chrge dq locted t distnce r from P. Then, tret one chrge element s point chrge, such tht the potentil t P due to the element is dv k e dq/r. Obtin the totl potentil t P by integrting dv over the entire chrge distribution. In performing the integrtion for most problems, you must express dq nd r in terms of single vrible. To simplify the integrtion, consider the geometry involved in the problem crefully. Review Exmples 25.5 through 25.7 for guidnce. Another method tht you cn use to obtin the electric potentil due to finite continuous chrge distribution is to strt with the definition of potentil difference given by Eqution If you know or cn esily obtin E (from Guss s lw), then you cn evlute the line integrl of E ds. An exmple of this method is given in Exmple Once you know the electric potentil t point, you cn obtin the electric field t tht point by remembering tht the electric field component in specified direction is equl to the negtive of the derivtive of the electric potentil in tht direction. Exmple 25.4 illustrtes this procedure. QUESTIONS 1. Distinguish between electric potentil nd electric potentil energy. 2. A negtive chrge moves in the direction of uniform electric field. Does the potentil energy of the chrge increse or decrese? Does it move to position of higher or lower potentil? 3. Give physicl explntion of the fct tht the potentil energy of pir of like chrges is positive wheres the potentil energy of pir of unlike chrges is negtive. 4. A uniform electric field is prllel to the x xis. In wht direction cn chrge be displced in this field without ny externl work being done on the chrge? 5. Explin why equipotentil surfces re lwys perpendiculr to electric field lines. 6. Describe the equipotentil surfces for () n infinite line of chrge nd (b) uniformly chrged sphere. 7. Explin why, under sttic conditions, ll points in conductor must be t the sme electric potentil. 8. The electric field inside hollow, uniformly chrged sphere is zero. Does this imply tht the potentil is zero inside the sphere? Explin. 9. The potentil of point chrge is defined to be zero t n infinite distnce. Why cn we not define the potentil of n infinite line of chrge to be zero t r? 10. Two chrged conducting spheres of different rdii re connected by conducting wire, s shown in Figure Which sphere hs the greter chrge density? 11. Wht determines the mximum potentil to which the dome of Vn de Grff genertor cn be rised? 12. Explin the origin of the glow sometimes observed round the cbles of high-voltge power line. 13. Why is it importnt to void shrp edges or points on conductors used in high-voltge equipment? 14. How would you shield n electronic circuit or lbortory from stry electric fields? Why does this work? 15. Why is it reltively sfe to sty in n utomobile with metl body during severe thunderstorm? 16. Wlking cross crpet nd then touching someone cn result in shock. Explin why this occurs.

90 Problems 795 PROBLEMS 1, 2, 3 = strightforwrd, intermedite, chllenging = full solution vilble in the Student Solutions Mnul nd Study Guide WEB = solution posted t = Computer useful in solving problem = Interctive Physics = pired numericl/symbolic problems Section 25.1 Potentil Difference nd Electric Potentil 1. How much work is done (by bttery, genertor, or some other source of electricl energy) in moving Avogdro s number of electrons from n initil point where the electric potentil is 9.00 V to point where the potentil is 5.00 V? (The potentil in ech cse is mesured reltive to common reference point.) 2. An ion ccelerted through potentil difference of 115 V experiences n increse in kinetic energy of J. Clculte the chrge on the ion. 3. () Clculte the speed of proton tht is ccelerted from rest through potentil difference of 120 V. (b) Clculte the speed of n electron tht is ccelerted through the sme potentil difference. 4. Review Problem. Through wht potentil difference would n electron need to be ccelerted for it to chieve speed of 40.0% of the speed of light, strting from rest? The speed of light is c m/s; review Section Wht potentil difference is needed to stop n electron hving n initil speed of m/s? Section 25.2 Potentil Differences in Uniform Electric Field 6. A uniform electric field of mgnitude 250 V/m is directed in the positive x direction. A 12.0-C chrge moves from the origin to the point (x, y) (20.0 cm, 50.0 cm). () Wht ws the chnge in the potentil energy of this chrge? (b) Through wht potentil difference did the chrge move? 7. The difference in potentil between the ccelerting pltes of TV set is bout V. If the distnce between these pltes is 1.50 cm, find the mgnitude of the uniform electric field in this region. 8. Suppose n electron is relesed from rest in uniform electric field whose mgnitude is V/m. () Through wht potentil difference will it hve pssed fter moving 1.00 cm? (b) How fst will the electron be moving fter it hs trveled 1.00 cm? 9. An electron moving prllel to the x xis hs n initil speed of m/s t the origin. Its speed is reduced to m/s t the point x 2.00 cm. Clculte the potentil difference between the origin nd tht point. Which point is t the higher potentil? 10. A uniform electric field of mgnitude 325 V/m is directed in the negtive y direction s shown in Figure P The coordintes of point A re ( 0.200, 0.300) m, nd those of point B re (0.400, 0.500) m. Clculte the potentil difference V B V A, using the blue pth. WEB 11. A 4.00-kg block crrying chrge Q 50.0 C is connected to spring for which k 100 N/m. The block lies on frictionless horizontl trck, nd the system is immersed in uniform electric field of mgnitude E V/m, directed s shown in Figure P If the block is relesed from rest when the spring is unstretched (t x 0), () by wht mximum mount does the spring expnd? (b) Wht is the equilibrium position of the block? (c) Show tht the block s motion is simple hrmonic, nd determine its period. (d) Repet prt () if the coefficient of kinetic friction between block nd surfce is A block hving mss m nd chrge Q is connected to spring hving constnt k. The block lies on frictionless horizontl trck, nd the system is immersed in uniform electric field of mgnitude E, directed s shown in Figure P If the block is relesed from rest when the spring is unstretched (t x 0), () by wht mximum mount does the spring expnd? (b) Wht is the equilibrium position of the block? (c) Show tht the block s motion is simple hrmonic, nd determine its period.(d) Repet prt () if the coefficient of kinetic friction between block nd surfce is k. k A y E Figure P25.10 m, Q x = 0 Figure P25.11 Problems 11 nd 12. E B x

91 796 CHAPTER 25 Electric Potentil 13. On plnet Tehr, the ccelertion due to grvity is the sme s tht on Erth but there is lso strong downwrd electric field with the field being uniform close to the plnet s surfce. A 2.00-kg bll hving chrge of 5.00 C is thrown upwrd t speed of 20.1 m/s nd it hits the ground fter n intervl of 4.10 s. Wht is the potentil difference between the strting point nd the top point of the trjectory? 14. An insulting rod hving liner chrge density 40.0 C/m nd liner mss density kg/m is relesed from rest in uniform electric field E 100 V/m directed perpendiculr to the rod (Fig. P25.14). () Determine the speed of the rod fter it hs trveled 2.00 m. (b) How does your nswer to prt () chnge if the electric field is not perpendiculr to the rod? Explin. string mkes n ngle 60.0 with uniform electric field of mgnitude E 300 V/m. Determine the speed of the prticle when the string is prllel to the electric field (point in Fig. P25.15). Section 25.3 Electric Potentil nd Potentil Energy Due to Point Chrges Note: Unless stted otherwise, ssume reference level of potentil V 0 t r. 16. () Find the potentil t distnce of 1.00 cm from proton. (b) Wht is the potentil difference between two points tht re 1.00 cm nd 2.00 cm from proton? (c) Repet prts () nd (b) for n electron. 17. Given two 2.00-C chrges, s shown in Figure P25.17, nd positive test chrge q C t the origin, () wht is the net force exerted on q by the two 2.00-C chrges? (b) Wht is the electric field t the origin due to the two 2.00-C chrges? (c) Wht is the electric potentil t the origin due to the two 2.00-C chrges? E E y 2.00 µ C q 2.00 µ C x x = m 0 x = m Figure P A prticle hving chrge q 2.00 C nd mss m kg is connected to string tht is L 1.50 m long nd is tied to the pivot point P in Figure P The prticle, string, nd pivot point ll lie on horizontl tble. The prticle is relesed from rest when the P θ L λ,, µ Figure P25.14 m q Top View Figure P25.15 E 18. A chrge q is t the origin. A chrge 2q is t x 2.00 m on the x xis. For wht finite vlue(s) of x is () the electric field zero? (b) the electric potentil zero? 19. The Bohr model of the hydrogen tom sttes tht the single electron cn exist only in certin llowed orbits round the proton. The rdius of ech Bohr orbit is r n 2 ( nm) where n 1, 2, 3,... Clculte the electric potentil energy of hydrogen tom when the electron is in the () first llowed orbit, n 1; (b) second llowed orbit, n 2; nd (c) when the electron hs escped from the tom (r ). Express your nswers in electron volts. 20. Two point chrges Q nc nd Q nc re seprted by 35.0 cm. () Wht is the potentil energy of the pir? Wht is the significnce of the lgebric sign of your nswer? (b) Wht is the electric potentil t point midwy between the chrges? 21. The three chrges in Figure P25.21 re t the vertices of n isosceles tringle. Clculte the electric potentil t the midpoint of the bse, tking q 7.00 C. 22. Compre this problem with Problem 55 in Chpter 23. Four identicl point chrges (q 10.0 C) re locted on the corners of rectngle, s shown in Figure P The dimensions of the rectngle re L 60.0 cm nd W 15.0 cm. Clculte the electric potentil energy of the chrge t the lower left corner due to the other three chrges.

92 Problems 797 q 4.00 cm q q 2.00 cm Figure P25.21 collide? (Hint: Consider conservtion of energy nd conservtion of liner momentum.) (b) If the spheres were conductors, would the speeds be greter or less thn those clculted in prt ()? 29. A smll sphericl object crries chrge of 8.00 nc. At wht distnce from the center of the object is the potentil equl to 100 V? 50.0 V? 25.0 V? Is the spcing of the equipotentils proportionl to the chnge in potentil? 30. Two point chrges of equl mgnitude re locted long the y xis equl distnces bove nd below the x xis, s shown in Figure P () Plot grph of the potentil t points long the x xis over the intervl 3 x 3. You should plot the potentil in units of k e Q /. (b) Let the chrge locted t be negtive nd plot the potentil long the y xis over the intervl 4 y 4. y WEB 23. Show tht the mount of work required to ssemble four identicl point chrges of mgnitude Q t the corners of squre of side s is 5.41k e Q 2 /s. 24. Compre this problem with Problem 18 in Chpter 23. Two point chrges ech of mgnitude 2.00 C re locted on the x xis. One is t x 1.00 m, nd the other is t x 1.00 m. () Determine the electric potentil on the y xis t y m. (b) Clculte the electric potentil energy of third chrge, of 3.00 C, plced on the y xis t y m. 25. Compre this problem with Problem 22 in Chpter 23. Five equl negtive point chrges q re plced symmetriclly round circle of rdius R. Clculte the electric potentil t the center of the circle. 26. Compre this problem with Problem 17 in Chpter 23. Three equl positive chrges q re t the corners of n equilterl tringle of side, s shown in Figure P () At wht point, if ny, in the plne of the chrges is the electric potentil zero? (b) Wht is the electric potentil t the point P due to the two chrges t the bse of the tringle? 27. Review Problem. Two insulting spheres hving rdii cm nd cm, msses kg nd kg, nd chrges 2.00 C nd 3.00 C re relesed from rest when their centers re seprted by 1.00 m. () How fst will ech be moving when they collide? (Hint: Consider conservtion of energy nd liner momentum.) (b) If the spheres were conductors would the speeds be lrger or smller thn those clculted in prt ()? Explin. 28. Review Problem. Two insulting spheres hving rdii r 1 nd r 2, msses m 1 nd m 2, nd chrges q 1 nd q 2 re relesed from rest when their centers re seprted by distnce d. () How fst is ech moving when they Q >O 31. In Rutherford s fmous scttering experiments tht led to the plnetry model of the tom, lph prticles (chrge 2e, mss kg) were fired t gold nucleus (chrge 79e). An lph prticle, initilly very fr from the gold nucleus, is fired with velocity of m/s directly towrd the center of the nucleus. How close does the lph prticle get to this center before turning round? Assume the gold nucleus remins sttionry. 32. An electron strts from rest 3.00 cm from the center of uniformly chrged insulting sphere of rdius 2.00 cm nd totl chrge 1.00 nc. Wht is the speed of the electron when it reches the surfce of the sphere? 33. Clculte the energy required to ssemble the rry of chrges shown in Figure P25.33, where m, b m, nd q 6.00 C. 34. Four identicl prticles ech hve chrge q nd mss m. They re relesed from rest t the vertices of squre of side L. How fst is ech chrge moving when their distnce from the center of the squre doubles? Q Figure P25.30 x

93 798 CHAPTER 25 Electric Potentil WEB q 2q 2q 35. How much work is required to ssemble eight identicl point chrges, ech of mgnitude q, t the corners of cube of side s? Section 25.4 Obtining the Vlue of the Electric Field from the Electric Potentil 36. The potentil in region between x 0 nd x 6.00 m is V bx where 10.0 V nd b 7.00 V/m. Determine () the potentil t x 0, 3.00 m, nd 6.00 m nd (b) the mgnitude nd direction of the electric field t x 0, 3.00 m, nd 6.00 m. 37. Over certin region of spce, the electric potentil is V 5x 3x 2 y 2yz 2. Find the expressions for the x, y, nd z components of the electric field over this region. Wht is the mgnitude of the field t the point P, which hs coordintes (1, 0, 2) m? 38. The electric potentil inside chrged sphericl conductor of rdius R is given by V k e Q /R nd outside the conductor is given by V k e Q /r. Using E r dv/dr, derive the electric field () inside nd (b) outside this chrge distribution. 39. It is shown in Exmple 25.7 tht the potentil t point P distnce bove one end of uniformly chrged rod of length lying long the x xis is V k eq ln!2 2 Use this result to derive n expression for the y component of the electric field t P. (Hint: Replce with y.) 40. When n unchrged conducting sphere of rdius is plced t the origin of n xyz coordinte system tht lies in n initilly uniform electric field E E 0 k, the resulting electric potentil is V(x, y, z) V 0 E 0 z b Figure P25.33 E 0 3 z (x 2 y 2 z 2 ) 3/2 for points outside the sphere, where V 0 is the (constnt) electric potentil on the conductor. Use this eqution to determine the x, y, nd z components of the resulting electric field. 3q Section 25.5 Electric Potentil Due to Continuous Chrge Distributions 41. Consider ring of rdius R with the totl chrge Q spred uniformly over its perimeter. Wht is the potentil difference between the point t the center of the ring nd point on its xis distnce 2R from the center? 42. Compre this problem with Problem 33 in Chpter 23. A uniformly chrged insulting rod of length 14.0 cm is bent into the shpe of semicircle, s shown in Figure P If the rod hs totl chrge of 7.50 C, find the electric potentil t O, the center of the semicircle. 43. A rod of length L (Fig. P25.43) lies long the x xis with its left end t the origin nd hs nonuniform chrge density x (where is positive constnt). () Wht re the units of? (b) Clculte the electric potentil t A. A d 44. For the rrngement described in the previous problem, clculte the electric potentil t point B tht lies on the perpendiculr bisector of the rod distnce b bove the x xis. 45. Clculte the electric potentil t point P on the xis of the nnulus shown in Figure P25.45, which hs uniform chrge density. y Figure P25.43 Problems 43 nd 44. b 46. A wire of finite length tht hs uniform liner chrge density is bent into the shpe shown in Figure P Find the electric potentil t point O. B L x b Figure P25.45 P x

94 Problems 799 WEB 2R Section 25.6 Electric Potentil Due to Chrged Conductor 47. How mny electrons should be removed from n initilly unchrged sphericl conductor of rdius m to produce potentil of 7.50 kv t the surfce? 48. Two chrged sphericl conductors re connected by long conducting wire, nd chrge of 20.0 C is plced on the combintion. () If one sphere hs rdius of 4.00 cm nd the other hs rdius of 6.00 cm, wht is the electric field ner the surfce of ech sphere? (b) Wht is the electric potentil of ech sphere? 49. A sphericl conductor hs rdius of 14.0 cm nd chrge of 26.0 C. Clculte the electric field nd the electric potentil t () r 10.0 cm, (b) r 20.0 cm, nd (c) r 14.0 cm from the center. 50. Two concentric sphericl conducting shells of rdii m nd b m re connected by thin wire, s shown in Figure P If totl chrge Q 10.0 C is plced on the system, how much chrge settles on ech sphere? (Optionl) Section 25.7 (Optionl) Section 25.8 q 2 b q 1 O The Millikn Oil-Drop Experiment Applictions of Electrosttics 51. Consider Vn de Grff genertor with 30.0-cmdimeter dome operting in dry ir. () Wht is the mximum potentil of the dome? (b) Wht is the mximum chrge on the dome? 52. The sphericl dome of Vn de Grff genertor cn be rised to mximum potentil of 600 kv; then dditionl chrge leks off in sprks, by producing brekdown of the surrounding dry ir. Determine () the chrge on the dome nd (b) the rdius of the dome. R Figure P25.46 Figure P25.50 Wire 2R ADDITIONAL PROBLEMS 53. The liquid-drop model of the nucleus suggests tht high-energy oscilltions of certin nuclei cn split the nucleus into two unequl frgments plus few neutrons. The frgments cquire kinetic energy from their mutul Coulomb repulsion. Clculte the electric potentil energy (in electron volts) of two sphericl frgments from urnium nucleus hving the following chrges nd rdii: 38e nd m; 54e nd m. Assume tht the chrge is distributed uniformly throughout the volume of ech sphericl frgment nd tht their surfces re initilly in contct t rest. (The electrons surrounding the nucleus cn be neglected.) 54. On dry winter dy you scuff your lether-soled shoes cross crpet nd get shock when you extend the tip of one finger towrd metl doorknob. In drk room you see sprk perhps 5 mm long. Mke order-ofmgnitude estimtes of () your electric potentil nd (b) the chrge on your body before you touch the doorknob. Explin your resoning. 55. The chrge distribution shown in Figure P25.55 is referred to s liner qudrupole. () Show tht the potentil t point on the x xis where x is V 2k eq 2 x 3 x 2 (b) Show tht the expression obtined in prt () when x W reduces to V 2k eq 2 x 3 Q 2Q Q (,0) 56. () Use the exct result from Problem 55 to find the electric field t ny point long the xis of the liner qudrupole for x. (b) Evlute E t x 3 if 2.00 mm nd Q 3.00 C. 57. At certin distnce from point chrge, the mgnitude of the electric field is 500 V/m nd the electric potentil is 3.00 kv. () Wht is the distnce to the chrge? (b) Wht is the mgnitude of the chrge? 58. An electron is relesed from rest on the xis of uniform positively chrged ring, m from the ring s y Qudrupole Figure P25.55 (,0) x

95 800 CHAPTER 25 Electric Potentil center. If the liner chrge density of the ring is C/m nd the rdius of the ring is m, how fst will the electron be moving when it reches the center of the ring? 59. () Consider uniformly chrged cylindricl shell hving totl chrge Q, rdius R, nd height h. Determine the electrosttic potentil t point distnce d from the right side of the cylinder, s shown in Figure P (Hint: Use the result of Exmple 25.5 by treting the cylinder s collection of ring chrges.) (b) Use the result of Exmple 25.6 to solve the sme problem for solid cylinder. r b Anode Cthode r λ λ λ Figure P Two prllel pltes hving chrges of equl mgnitude but opposite sign re seprted by 12.0 cm. Ech plte hs surfce chrge density of 36.0 nc/m 2. A proton is relesed from rest t the positive plte. Determine () the potentil difference between the pltes, (b) the energy of the proton when it reches the negtive plte, (c) the speed of the proton just before it strikes the negtive plte, (d) the ccelertion of the proton, nd (e) the force on the proton. (f) From the force, find the mgnitude of the electric field nd show tht it is equl to tht found from the chrge densities on the pltes. 61. Clculte the work tht must be done to chrge sphericl shell of rdius R to totl chrge Q. 62. A GeigerMüller counter is rdition detector tht essentilly consists of hollow cylinder (the cthode) of inner rdius r nd coxil cylindricl wire (the node) of rdius r b (Fig. P25.62). The chrge per unit length on the node is, while the chrge per unit length on the cthode is. () Show tht the mgnitude of the potentil difference between the wire nd the cylinder in the sensitive region of the detector is (b) Show tht the mgnitude of the electric field over tht region is given by E h V 2k e ln r r b V ln(r /r b ) 1 r where r is the distnce from the center of the node to the point where the field is to be clculted. R Figure P25.59 d WEB 63. From Guss s lw, the electric field set up by uniform line of chrge is E 20r rˆ where rˆ is unit vector pointing rdilly wy from the line nd is the chrge per unit length long the line. Derive n expression for the potentil difference between r r 1 nd r r A point chrge q is locted t x R, nd point chrge 2q is locted t the origin. Prove tht the equipotentil surfce tht hs zero potentil is sphere centered t ( 4R/3, 0, 0) nd hving rdius r 2R/ Consider two thin, conducting, sphericl shells s shown in cross-section in Figure P The inner shell hs rdius r cm nd chrge of 10.0 nc. The outer shell hs rdius r cm nd chrge of 15.0 nc. Find () the electric field E nd (b) the electric potentil V in regions A, B, nd C, with V 0 t r. C B A 66. The x xis is the symmetry xis of uniformly chrged ring of rdius R nd chrge Q (Fig. P25.66). A point chrge Q of mss M is locted t the center of the ring. When it is displced slightly, the point chrge cceler- r 1 r 2 Figure P25.65

96 Problems 801 tes long the x xis to infinity. Show tht the ultimte speed of the point chrge is v 2k e Q 2 MR 1/2 y r 1 P E r E θ Q q r R Uniformly chrged ring Figure P25.66 Q v x θ q r 2 x 67. An infinite sheet of chrge tht hs surfce chrge density of 25.0 nc/m 2 lies in the yz plne, psses through the origin, nd is t potentil of 1.00 kv t the point y 0, z 0. A long wire hving liner chrge density of 80.0 nc/m lies prllel to the y xis nd intersects the x xis t x 3.00 m. () Determine, s function of x, the potentil long the x xis between wire nd sheet. (b) Wht is the potentil energy of 2.00-nC chrge plced t x m? 68. The thin, uniformly chrged rod shown in Figure P25.68 hs liner chrge density. Find n expression for the electric potentil t P. P y Figure P25.69 (b) For the dipole rrngement shown, express V in terms of crtesin coordintes using r (x 2 y 2 ) 1/2 nd cos y (x 2 y 2 ) 1/2 Using these results nd tking r W, clculte the field components E x nd E y. 70. Figure P25.70 shows severl equipotentil lines ech lbeled by its potentil in volts. The distnce between the lines of the squre grid represents 1.00 cm. () Is the mgnitude of the field bigger t A or t B? Why? (b) Wht is E t B? (c) Represent wht the field looks like by drwing t lest eight field lines. b A L Figure P25.68 x A dipole is locted long the y xis s shown in Figure P () At point P, which is fr from the dipole (r W ), the electric potentil is V k e p cos r 2 where p 2q. Clculte the rdil component E r nd the perpendiculr component E of the ssocited electric field. Note tht Do these results seem resonble for 90 nd 0? for r 0? E (1/r)(V/). Figure P A disk of rdius R hs nonuniform surfce chrge density Cr, where C is constnt nd r is mesured from the center of the disk (Fig. P25.71). Find (by direct integrtion) the potentil t P. B

97 802 CHAPTER 25 Electric Potentil R Figure P A solid sphere of rdius R hs uniform chrge density nd totl chrge Q. Derive n expression for its totl x P electric potentil energy. (Hint: Imgine tht the sphere is constructed by dding successive lyers of concentric shells of chrge dq (4r 2 dr) nd use du V dq.) 73. The results of Problem 62 pply lso to n electrosttic precipittor (see Figs nd P25.62). An pplied voltge V V V b 50.0 kv is to produce n electric field of mgnitude 5.50 MV/m t the surfce of the centrl wire. The outer cylindricl wll hs uniform rdius r m. () Wht should be the rdius r b of the centrl wire? You will need to solve trnscendentl eqution. (b) Wht is the mgnitude of the electric field t the outer wll? ANSWERS TO QUICK QUIZZES 25.1 We do if the electric field is uniform. (This is precisely wht we do in the next section.) In generl, however, n electric field chnges from one plce to nother B : C, C : D, A : B, D : E. Moving from B to C decreses the electric potentil by 2 V, so the electric field performs 2 J of work on ech coulomb of chrge tht moves. Moving from C to D decreses the electric potentil by 1 V, so 1 J of work is done by the field. It tkes no work to move the chrge from A to B becuse the electric potentil does not chnge. Moving from D to E increses the electric potentil by 1 V, nd thus the field does 1 J of work, just s rising mss to higher elevtion cuses the grvittionl field to do negtive work on the mss The electric potentil decreses in inverse proportion to the rdius (see Eq ). The electric field mgnitude decreses s the reciprocl of the rdius squred (see Eq. 23.4). Becuse the surfce re increses s r 2 while the electric field mgnitude decreses s 1/r 2, the electric flux through the surfce remins constnt (see Eq. 24.1) () Yes. Consider four equl chrges plced t the corners of squre. The electric potentil grph for this sitution is shown in the figure. At the center of the squre, the electric field is zero becuse the individul fields from the four chrges cncel, but the potentil is not zero. This is lso the sitution inside chrged conductor. (b) Yes gin. In Figure 25.8, for instnce, the electric potentil is zero t the center of the dipole, but the mgnitude of the field t tht point is not zero. (The two chrges in dipole re by definition of opposite sign; thus, the electric field lines creted by the two chrges extend from the positive to the negtive chrge nd do not cncel nywhere.) This is the sitution we presented in Exmple 25.4c, in which the equtions we obtined give V 0 nd E x 0. Electric potentil (V) x y

98

99 2.2 This is the Nerest One Hed 803 P U Z Z L E R Mny electronic components crry wrning lbel like this one. Wht is there inside these devices tht mkes them so dngerous? Why wouldn t you be sfe if you unplugged the equipment before opening the cse? (George Semple) Cpcitnce nd Dielectrics c h p t e r Chpter Outline 26.1 Definition of Cpcitnce 26.2 Clculting Cpcitnce 26.3 Combintions of Cpcitors 26.4 Energy Stored in Chrged Cpcitor 26.5 Cpcitors with Dielectrics 26.6 (Optionl) Electric Dipole in n Electric Field 26.7 (Optionl) An Atomic Description of Dielectrics 803

100 804 CHAPTER 26 Cpcitnce nd Dielectrics In this chpter, we discuss cpcitors devices tht store electric chrge. Cpcitors re commonly used in vriety of electric circuits. For instnce, they re used to tune the frequency of rdio receivers, s filters in power supplies, to eliminte sprking in utomobile ignition systems, nd s energy-storing devices in electronic flsh units. A cpcitor consists of two conductors seprted by n insultor. We shll see tht the cpcitnce of given cpcitor depends on its geometry nd on the mteril clled dielectric tht seprtes the conductors DEFINITION OF CAPACITANCE Consider two conductors crrying chrges of equl mgnitude but of opposite sign, s shown in Figure Such combintion of two conductors is clled cpcitor. The conductors re clled pltes. A potentil difference V exists between the conductors due to the presence of the chrges. Becuse the unit of potentil difference is the volt, potentil difference is often clled voltge. We shll use this term to describe the potentil difference cross circuit element or between two points in spce. Wht determines how much chrge is on the pltes of cpcitor for given voltge? In other words, wht is the cpcity of the device for storing chrge t prticulr vlue of V? Experiments show tht the quntity of chrge Q on cpcitor 1 is linerly proportionl to the potentil difference between the conductors; tht is, Q V. The proportionlity constnt depends on the shpe nd seprtion of the conductors. 2 We cn write this reltionship s Q C V if we define cpcitnce s follows: Definition of cpcitnce The cpcitnce C of cpcitor is the rtio of the mgnitude of the chrge on either conductor to the mgnitude of the potentil difference between them: C Q V (26.1) Q Q Note tht by definition cpcitnce is lwys positive quntity. Furthermore, the potentil difference V is lwys expressed in Eqution 26.1 s positive quntity. Becuse the potentil difference increses linerly with the stored chrge, the rtio Q /V is constnt for given cpcitor. Therefore, cpcitnce is mesure of cpcitor s bility to store chrge nd electric potentil energy. From Eqution 26.1, we see tht cpcitnce hs SI units of coulombs per volt. The SI unit of cpcitnce is the frd (F), which ws nmed in honor of Michel Frdy: 1 F 1 C/V The frd is very lrge unit of cpcitnce. In prctice, typicl devices hve cpcitnces rnging from microfrds (10 6 F) to picofrds (10 12 F). For prcticl purposes, cpcitors often re lbeled mf for microfrds nd mmf for micromicrofrds or, equivlently, pf for picofrds. Figure 26.1 A cpcitor consists of two conductors crrying chrges of equl mgnitude but opposite sign. 1 Although the totl chrge on the cpcitor is zero (becuse there is s much excess positive chrge on one conductor s there is excess negtive chrge on the other), it is common prctice to refer to the mgnitude of the chrge on either conductor s the chrge on the cpcitor. 2 The proportionlity between V nd Q cn be proved from Coulomb s lw or by experiment.

101 26.2 Clculting Cpcitnce 805 Q Q A collection of cpcitors used in vriety of pplictions. Let us consider cpcitor formed from pir of prllel pltes, s shown in Figure Ech plte is connected to one terminl of bttery (not shown in Fig. 26.2), which cts s source of potentil difference. If the cpcitor is initilly unchrged, the bttery estblishes n electric field in the connecting wires when the connections re mde. Let us focus on the plte connected to the negtive terminl of the bttery. The electric field pplies force on electrons in the wire just outside this plte; this force cuses the electrons to move onto the plte. This movement continues until the plte, the wire, nd the terminl re ll t the sme electric potentil. Once this equilibrium point is ttined, potentil difference no longer exists between the terminl nd the plte, nd s result no electric field is present in the wire, nd the movement of electrons stops. The plte now crries negtive chrge. A similr process occurs t the other cpcitor plte, with electrons moving from the plte to the wire, leving the plte positively chrged. In this finl configurtion, the potentil difference cross the cpcitor pltes is the sme s tht between the terminls of the bttery. Suppose tht we hve cpcitor rted t 4 pf. This rting mens tht the cpcitor cn store 4 pc of chrge for ech volt of potentil difference between the two conductors. If 9-V bttery is connected cross this cpcitor, one of the conductors ends up with net chrge of 36 pc nd the other ends up with net chrge of 36 pc. Figure 26.2 d Are = A A prllel-plte cpcitor consists of two prllel conducting pltes, ech of re A, seprted by distnce d. When the cpcitor is chrged, the pltes crry equl mounts of chrge. One plte crries positive chrge, nd the other crries negtive chrge CALCULATING CAPACITANCE We cn clculte the cpcitnce of pir of oppositely chrged conductors in the following mnner: We ssume chrge of mgnitude Q, nd we clculte the potentil difference using the techniques described in the preceding chpter. We then use the expression C Q /V to evlute the cpcitnce. As we might expect, we cn perform this clcultion reltively esily if the geometry of the cpcitor is simple. We cn clculte the cpcitnce of n isolted sphericl conductor of rdius R nd chrge Q if we ssume tht the second conductor mking up the cpcitor is concentric hollow sphere of infinite rdius. The electric potentil of the sphere of rdius R is simply k e Q /R, nd setting V 0 t infinity s usul, we hve C Q (26.2) V Q k e Q /R R 40R k e This expression shows tht the cpcitnce of n isolted chrged sphere is proportionl to its rdius nd is independent of both the chrge on the sphere nd the potentil difference. QuickLb Roll some socks into blls nd stuff them into shoebox. Wht determines how mny socks fit in the box? Relte how hrd you push on the socks to V for cpcitor. How does the size of the box influence its sock cpcity?

102 806 CHAPTER 26 Cpcitnce nd Dielectrics The cpcitnce of pir of conductors depends on the geometry of the conductors. Let us illustrte this with three fmilir geometries, nmely, prllel pltes, concentric cylinders, nd concentric spheres. In these exmples, we ssume tht the chrged conductors re seprted by vcuum. The effect of dielectric mteril plced between the conductors is treted in Section Prllel-Plte Cpcitors Two prllel metllic pltes of equl re A re seprted by distnce d, s shown in Figure One plte crries chrge Q,nd the other crries chrge Q. Let us consider how the geometry of these conductors influences the cpcity of the combintion to store chrge. Recll tht chrges of like sign repel one nother. As cpcitor is being chrged by bttery, electrons flow into the negtive plte nd out of the positive plte. If the cpcitor pltes re lrge, the ccumulted chrges re ble to distribute themselves over substntil re, nd the mount of chrge tht cn be stored on plte for given potentil difference increses s the plte re is incresed. Thus, we expect the cpcitnce to be proportionl to the plte re A. Now let us consider the region tht seprtes the pltes. If the bttery hs constnt potentil difference between its terminls, then the electric field between the pltes must increse s d is decresed. Let us imgine tht we move the pltes closer together nd consider the sitution before ny chrges hve hd chnce to move in response to this chnge. Becuse no chrges hve moved, the electric field between the pltes hs the sme vlue but extends over shorter distnce. Thus, the mgnitude of the potentil difference between the pltes V Ed (Eq. 25.6) is now smller. The difference between this new cpcitor voltge nd the terminl voltge of the bttery now exists s potentil difference cross the wires connecting the bttery to the cpcitor. This potentil difference results in n electric field in the wires tht drives more chrge onto the pltes, incresing the potentil difference between the pltes. When the potentil difference between the pltes gin mtches tht of the bttery, the potentil difference cross the wires flls bck to zero, nd the flow of chrge stops. Thus, moving the pltes closer together cuses the chrge on the cpcitor to increse. If d is incresed, the chrge decreses. As result, we expect the device s cpcitnce to be inversely proportionl to d. Q Q Figure 26.3 () () The electric field between the pltes of prllel-plte cpcitor is uniform ner the center but nonuniform ner the edges. (b) Electric field pttern of two oppositely chrged conducting prllel pltes. Smll pieces of thred on n oil surfce lign with the electric field. (b)

103 26.2 Clculting Cpcitnce 807 We cn verify these physicl rguments with the following derivtion. The surfce chrge density on either plte is If the pltes re very close together (in comprison with their length nd width), we cn ssume tht the electric field is uniform between the pltes nd is zero elsewhere. According to the lst prgrph of Exmple 24.8, the vlue of the electric field between the pltes is E Becuse the field between the pltes is uniform, the mgnitude of the potentil difference between the pltes equls Ed (see Eq. 25.6); therefore, V Ed Qd 0A Substituting this result into Eqution 26.1, we find tht the cpcitnce is 0 Q /A. Q 0A C Q V Q Qd/0A C 0A d (26.3) Tht is, the cpcitnce of prllel-plte cpcitor is proportionl to the re of its pltes nd inversely proportionl to the plte seprtion, just s we expect from our conceptul rgument. A creful inspection of the electric field lines for prllel-plte cpcitor revels tht the field is uniform in the centrl region between the pltes, s shown in Figure However, the field is nonuniform t the edges of the pltes. Figure 26.3b is photogrph of the electric field pttern of prllel-plte cpcitor. Note the nonuniform nture of the electric field t the ends of the pltes. Such end effects cn be neglected if the plte seprtion is smll compred with the length of the pltes. Quick Quiz 26.1 Mny computer keybord buttons re constructed of cpcitors, s shown in Figure When key is pushed down, the soft insultor between the movble plte nd the fixed plte is compressed. When the key is pressed, the cpcitnce () increses, (b) decreses, or (c) chnges in wy tht we cnnot determine becuse the complicted electric circuit connected to the keybord button my cuse chnge in V. Key Movble plte Soft insultor Fixed plte Figure 26.4 B One type of computer keybord button. EXAMPLE 26.1 Prllel-Plte Cpcitor A prllel-plte cpcitor hs n re A m 2 nd plte seprtion d 1.00 mm. Find its cpcitnce F 1.77 pf Solution From Eqution 26.3, we find tht C A 0 d ( C 2 /Nm 2 ) m m Exercise 3.00 mm? Answer Wht is the cpcitnce for plte seprtion of pf.

104 808 CHAPTER 26 Cpcitnce nd Dielectrics Cylindricl nd Sphericl Cpcitors From the definition of cpcitnce, we cn, in principle, find the cpcitnce of ny geometric rrngement of conductors. The following exmples demonstrte the use of this definition to clculte the cpcitnce of the other fmilir geometries tht we mentioned: cylinders nd spheres. EXAMPLE 26.2 The Cylindricl Cpcitor A solid cylindricl conductor of rdius nd chrge Q is coxil with cylindricl shell of negligible thickness, rdius b, nd chrge Q (Fig. 26.5). Find the cpcitnce of this cylindricl cpcitor if its length is. Solution It is difficult to pply physicl rguments to this configurtion, lthough we cn resonbly expect the cpcitnce to be proportionl to the cylinder length for the sme reson tht prllel-plte cpcitnce is proportionl to plte re: Stored chrges hve more room in which to be distributed. If we ssume tht is much greter thn nd b, we cn neglect end effects. In this cse, the electric field is perpendiculr to the long xis of the cylinders nd is confined to the region between them (Fig. 26.5b). We must first clculte the potentil difference between the two cylinders, which is given in generl by where E is the electric field in the region r b. In Chpter 24, we showed using Guss s lw tht the mgnitude of the electric field of cylindricl chrge distribution hving liner chrge density is E r 2k e/r (Eq. 24.7). The sme result pplies here becuse, ccording to Guss s lw, the chrge on the outer cylinder does not contribute to the electric field inside it. Using this result nd noting from Figure 26.5b tht E is long r, we find tht V b V b E r dr 2k e b Substituting this result into Eqution 26.1 nd using the fct tht we obtin Q /, C Q V Q 2k e Q V b V b E ds ln b (26.4) where V is the mgnitude of the potentil difference, given dr r 2k e ln b 2k e ln b by V V b V 2k e ln (b/), positive quntity. As predicted, the cpcitnce is proportionl to the length of the cylinders. As we might expect, the cpcitnce lso depends on the rdii of the two cylindricl conductors. From Eqution 26.4, we see tht the cpcitnce per unit length of combintion of concentric cylindricl conductors is C 1 2k e ln b (26.5) An exmple of this type of geometric rrngement is coxil cble, which consists of two concentric cylindricl conductors seprted by n insultor. The cble crries electricl signls in the inner nd outer conductors. Such geometry is especilly useful for shielding the signls from ny possible externl influences. b Figure 26.5 () Gussin surfce Q () A cylindricl cpcitor consists of solid cylindricl conductor of rdius nd length surrounded by coxil cylindricl shell of rdius b. (b) End view. The dshed line represents the end of the cylindricl gussin surfce of rdius r nd length. r b (b) Q EXAMPLE 26.3 The Sphericl Cpcitor A sphericl cpcitor consists of sphericl conducting shell of rdius b nd chrge Q concentric with smller conducting sphere of rdius nd chrge Q (Fig. 26.6). Find the cpcitnce of this device. Solution As we showed in Chpter 24, the field outside sphericlly symmetric chrge distribution is rdil nd given by the expression k e Q /r 2. In this cse, this result pplies to the field between the spheres ( r b). From

105 26.3 Combintions of Cpcitors 809 Guss s lw we see tht only the inner sphere contributes to this field. Thus, the potentil difference between the spheres is Q V b V b E r dr k e Q b k e Q 1 b 1 dr r 2 k eq 1 r b b Q The mgnitude of the potentil difference is V V b V k e Q (b ) b Substituting this vlue for V into Eqution 26.1, we obtin Figure 26.6 A sphericl cpcitor consists of n inner sphere of rdius surrounded by concentric sphericl shell of rdius b. The electric field between the spheres is directed rdilly outwrd when the inner sphere is positively chrged. C Q V b k e (b ) (26.6) Exercise Show tht s the rdius b of the outer sphere pproches infinity, the cpcitnce pproches the vlue /k e 40. Quick Quiz 26.2 Wht is the mgnitude of the electric field in the region outside the sphericl cpcitor described in Exmple 26.3? COMBINATIONS OF CAPACITORS Two or more cpcitors often re combined in electric circuits. We cn clculte the equivlent cpcitnce of certin combintions using methods described in this section. The circuit symbols for cpcitors nd btteries, s well s the color codes used for them in this text, re given in Figure The symbol for the cpcitor reflects the geometry of the most common model for cpcitor pir of prllel pltes. The positive terminl of the bttery is t the higher potentil nd is represented in the circuit symbol by the longer verticl line. Prllel Combintion Two cpcitors connected s shown in Figure 26.8 re known s prllel combintion of cpcitors. Figure 26.8b shows circuit digrm for this combintion of cpcitors. The left pltes of the cpcitors re connected by conducting wire to the positive terminl of the bttery nd re therefore both t the sme electric potentil s the positive terminl. Likewise, the right pltes re connected to the negtive terminl nd re therefore both t the sme potentil s the negtive terminl. Thus, the individul potentil differences cross cpcitors connected in prllel re ll the sme nd re equl to the potentil difference pplied cross the combintion. In circuit such s tht shown in Figure 26.8, the voltge pplied cross the combintion is the terminl voltge of the bttery. Situtions cn occur in which Figure 26.7 Cpcitor symbol Bttery symbol Switch symbol Circuit symbols for cpcitors, btteries, nd switches. Note tht cpcitors re in blue nd btteries nd switches re in red.

106 810 CHAPTER 26 Cpcitnce nd Dielectrics C 1 V 1 = V 2 = V C 1 C eq = C 1 C 2 Q 1 C 2 C 2 Q 2 V V V Figure 26.8 () () A prllel combintion of two cpcitors in n electric circuit in which the potentil difference cross the bttery terminls is V. (b) The circuit digrm for the prllel combintion. (c) The equivlent cpcitnce is C eq C 1 C 2. (b) (c) the prllel combintion is in circuit with other circuit elements; in such situtions, we must determine the potentil difference cross the combintion by nlyzing the entire circuit. When the cpcitors re first connected in the circuit shown in Figure 26.8, electrons re trnsferred between the wires nd the pltes; this trnsfer leves the left pltes positively chrged nd the right pltes negtively chrged. The energy source for this chrge trnsfer is the internl chemicl energy stored in the bttery, which is converted to electric potentil energy ssocited with the chrge seprtion. The flow of chrge ceses when the voltge cross the cpcitors is equl to tht cross the bttery terminls. The cpcitors rech their mximum chrge when the flow of chrge ceses. Let us cll the mximum chrges on the two cpcitors Q 1 nd Q 2. The totl chrge Q stored by the two cpcitors is Q Q 1 Q 2 (26.7) Tht is, the totl chrge on cpcitors connected in prllel is the sum of the chrges on the individul cpcitors. Becuse the voltges cross the cpcitors re the sme, the chrges tht they crry re Q 1 C 1 V Q 2 C 2 V Suppose tht we wish to replce these two cpcitors by one equivlent cpcitor hving cpcitnce C eq, s shown in Figure 26.8c. The effect this equivlent cpcitor hs on the circuit must be exctly the sme s the effect of the combintion of the two individul cpcitors. Tht is, the equivlent cpcitor must store Q units of chrge when connected to the bttery. We cn see from Figure 26.8c tht the voltge cross the equivlent cpcitor lso is V becuse the equivlent cpc-

107 26.3 Combintions of Cpcitors 811 itor is connected directly cross the bttery terminls. Thus, for the equivlent cpcitor, Q C eq V Substituting these three reltionships for chrge into Eqution 26.7, we hve C eq V C 1 V C 2 V C eq C 1 C 2 prllel combintion If we extend this tretment to three or more cpcitors connected in prllel, we find the equivlent cpcitnce to be C eq C 1 C 2 C 3 (prllel combintion) (26.8) Thus, the equivlent cpcitnce of prllel combintion of cpcitors is greter thn ny of the individul cpcitnces. This mkes sense becuse we re essentilly combining the res of ll the cpcitor pltes when we connect them with conducting wire. Series Combintion Two cpcitors connected s shown in Figure 26.9 re known s series combintion of cpcitors. The left plte of cpcitor 1 nd the right plte of cpcitor 2 re connected to the terminls of bttery. The other two pltes re connected to ech other nd to nothing else; hence, they form n isolted conductor tht is initilly unchrged nd must continue to hve zero net chrge. To nlyze this combintion, let us begin by considering the unchrged cpcitors nd follow wht hppens just fter bttery is connected to the circuit. When the bttery is con- C 1 V 1 V 2 C 2 C eq Q Q Q Q V V Figure 26.9 () () A series combintion of two cpcitors. The chrges on the two cpcitors re the sme. (b) The cpcitors replced by single equivlent cpcitor. The equivlent cpcitnce cn be clculted from the reltionship 1 C eq 1 C 1 1 C 2 (b)

108 812 CHAPTER 26 Cpcitnce nd Dielectrics nected, electrons re trnsferred out of the left plte of C 1 nd into the right plte of C 2. As this negtive chrge ccumultes on the right plte of C 2, n equivlent mount of negtive chrge is forced off the left plte of C 2, nd this left plte therefore hs n excess positive chrge. The negtive chrge leving the left plte of C 2 trvels through the connecting wire nd ccumultes on the right plte of C 1. As result, ll the right pltes end up with chrge Q, nd ll the left pltes end up with chrge Q. Thus, the chrges on cpcitors connected in series re the sme. From Figure 26.9, we see tht the voltge V cross the bttery terminls is split between the two cpcitors: V V 1 V 2 (26.9) where V 1 nd V 2 re the potentil differences cross cpcitors C 1 nd C 2, respectively. In generl, the totl potentil difference cross ny number of cpcitors connected in series is the sum of the potentil differences cross the individul cpcitors. Suppose tht n equivlent cpcitor hs the sme effect on the circuit s the series combintion. After it is fully chrged, the equivlent cpcitor must hve chrge of Q on its right plte nd chrge of Q on its left plte. Applying the definition of cpcitnce to the circuit in Figure 26.9b, we hve V Becuse we cn pply the expression Q C V to ech cpcitor shown in Figure 26.9, the potentil difference cross ech is Q C eq V 1 Q C 1 V 2 Q C 2 Substituting these expressions into Eqution 26.9 nd noting tht we hve Q C eq Q C 1 Q C 2 Cnceling Q, we rrive t the reltionship 1 C eq 1 C 1 1 C 2 series combintion V Q /C eq, When this nlysis is pplied to three or more cpcitors connected in series, the reltionship for the equivlent cpcitnce is 1 C eq 1 C 1 1 C 2 1 C 3 series combintion (26.10) This demonstrtes tht the equivlent cpcitnce of series combintion is lwys less thn ny individul cpcitnce in the combintion. EXAMPLE 26.4 Equivlent Cpcitnce Find the equivlent cpcitnce between nd b for the combintion of cpcitors shown in Figure All cpcitnces re in microfrds. Solution Using Equtions 26.8 nd 26.10, we reduce the combintion step by step s indicted in the figure. The 1.0-F nd 3.0-F cpcitors re in prllel nd combine c-

109 26.4 Energy Stored in Chrged Cpcitor 813 cording to the expression C eq C 1 C F. The 2.0-F nd 6.0-F cpcitors lso re in prllel nd hve n equivlent cpcitnce of 8.0 F. Thus, the upper brnch in Figure 26.10b consists of two 4.0-F cpcitors in series, which combine s follows: 1 C eq 1 C 1 1 C 2 C eq 1 1/2.0 F 2.0 F F F F The lower brnch in Figure 26.10b consists of two 8.0-F cpcitors in series, which combine to yield n equivlent cpcitnce of 4.0 F. Finlly, the 2.0-F nd 4.0-F cpcitors in Figure 26.10c re in prllel nd thus hve n equivlent cpcitnce of 6.0 F. Exercise Consider three cpcitors hving cpcitnces of 3.0 F, 6.0 F, nd 12 F. Find their equivlent cpcitnce when they re connected () in prllel nd (b) in series. Answer () 21 F; (b) 1.7 F b b b 6.0 b () Figure (b) To find the equivlent cpcitnce of the cpcitors in prt (), we reduce the vrious combintions in steps s indicted in prts (b), (c), nd (d), using the series nd prllel rules described in the text. (c) (d) ENERGYSTORED IN A CHARGED CAPACITOR Almost everyone who works with electronic equipment hs t some time verified tht cpcitor cn store energy. If the pltes of chrged cpcitor re connected by conductor, such s wire, chrge moves between the pltes nd the connecting wire until the cpcitor is unchrged. The dischrge cn often be observed s visible sprk. If you should ccidentlly touch the opposite pltes of chrged cpcitor, your fingers ct s pthwy for dischrge, nd the result is n electric shock. The degree of shock you receive depends on the cpcitnce nd on the voltge pplied to the cpcitor. Such shock could be ftl if high voltges re present, such s in the power supply of television set. Becuse the chrges cn be stored in cpcitor even when the set is turned off, unplugging the television does not mke it sfe to open the cse nd touch the components inside. Consider prllel-plte cpcitor tht is initilly unchrged, such tht the initil potentil difference cross the pltes is zero. Now imgine tht the cpcitor is connected to bttery nd develops mximum chrge Q. (We ssume tht the cpcitor is chrged slowly so tht the problem cn be considered s n electrosttic system.) When the cpcitor is connected to the bttery, electrons in the wire just outside the plte connected to the negtive terminl move into the plte to give it negtive chrge. Electrons in the plte connected to the positive terminl move out of the plte into the wire to give the plte positive chrge. Thus, chrges move only smll distnce in the wires. To clculte the energy of the cpcitor, we shll ssume different process one tht does not ctully occur but gives the sme finl result. We cn mke this

110 814 CHAPTER 26 Cpcitnce nd Dielectrics QuickLb Here s how to find out whether your clcultor hs cpcitor to protect vlues or progrms during bttery chnges: Store number in your clcultor s memory, remove the clcultor bttery for moment, nd then quickly replce it. Ws the number tht you stored preserved while the bttery ws out of the clcultor? (You my wnt to write down ny criticl numbers or progrms tht re stored in the clcultor before trying this!) ssumption becuse the energy in the finl configurtion does not depend on the ctul chrge-trnsfer process. We imgine tht we rech in nd grb smll mount of positive chrge on the plte connected to the negtive terminl nd pply force tht cuses this positive chrge to move over to the plte connected to the positive terminl. Thus, we do work on the chrge s we trnsfer it from one plte to the other. At first, no work is required to trnsfer smll mount of chrge dq from one plte to the other. 3 However, once this chrge hs been trnsferred, smll potentil difference exists between the pltes. Therefore, work must be done to move dditionl chrge through this potentil difference. As more nd more chrge is trnsferred from one plte to the other, the potentil difference increses in proportion, nd more work is required. Suppose tht q is the chrge on the cpcitor t some instnt during the chrging process. At the sme instnt, the potentil difference cross the cpcitor is V q/c. From Section 25.2, we know tht the work necessry to trnsfer n increment of chrge dq from the plte crrying chrge q to the plte crrying chrge q (which is t the higher electric potentil) is dw V dq q C dq Energy stored in chrged cpcitor This is illustrted in Figure The totl work required to chrge the cpcitor from q 0 to some finl chrge q Q is W Q q 0 C dq 1 Q q dq Q 2 C 0 2C The work done in chrging the cpcitor ppers s electric potentil energy U stored in the cpcitor. Therefore, we cn express the potentil energy stored in chrged cpcitor in the following forms: U Q 2 (26.11) 2C 1 2 Q V 1 2C(V )2 This result pplies to ny cpcitor, regrdless of its geometry. We see tht for given cpcitnce, the stored energy increses s the chrge increses nd s the potentil difference increses. In prctice, there is limit to the mximum energy V dq q Figure A plot of potentil difference versus chrge for cpcitor is stright line hving slope 1/C. The work required to move chrge dq through the potentil difference V cross the cpcitor pltes is given by the re of the shded rectngle. The totl work required to chrge the cpcitor to finl chrge Q is the tringulr re under the stright line, W 1 2 Q V. (Don t forget tht 1 V 1 J/C; hence, the unit for the re is the joule.) 3 We shll use lowercse q for the vrying chrge on the cpcitor while it is chrging, to distinguish it from uppercse Q, which is the totl chrge on the cpcitor fter it is completely chrged.

111 26.4 Energy Stored in Chrged Cpcitor 815 (or chrge) tht cn be stored becuse, t sufficiently gret vlue of V, dischrge ultimtely occurs between the pltes. For this reson, cpcitors re usully lbeled with mximum operting voltge. Quick Quiz 26.3 You hve three cpcitors nd bttery. How should you combine the cpcitors nd the bttery in one circuit so tht the cpcitors will store the mximum possible energy? We cn consider the energy stored in cpcitor s being stored in the electric field creted between the pltes s the cpcitor is chrged. This description is resonble in view of the fct tht the electric field is proportionl to the chrge on the cpcitor. For prllel-plte cpcitor, the potentil difference is relted to the electric field through the reltionship V Ed. Furthermore, its cpcitnce is C 0A/d (Eq. 26.3). Substituting these expressions into Eqution 26.11, we obtin U 1 0A (E 2 d 2 ) 1 (26.12) 2 d 2 (0Ad)E 2 Becuse the volume V (volume, not voltge!) occupied by the electric field is Ad, the energy per unit volume u E U/V U/Ad, known s the energy density, is Energy stored in prllel-plte cpcitor u E 1 20E 2 (26.13) Energy density in n electric field Although Eqution ws derived for prllel-plte cpcitor, the expression is generlly vlid. Tht is, the energy density in ny electric field is proportionl to the squre of the mgnitude of the electric field t given point. This bnk of cpcitors stores electricl energy for use in the prticle ccelertor t FermiLb, locted outside Chicgo. Becuse the electric utility compny cnnot provide lrge enough burst of energy to operte the equipment, these cpcitors re slowly chrged up, nd then the energy is rpidly dumped into the ccelertor. In this sense, the setup is much like fireprotection wter tnk on top of building. The tnk collects wter nd stores it for situtions in which lot of wter is needed in short time.

112 816 CHAPTER 26 Cpcitnce nd Dielectrics EXAMPLE 26.5 Rewiring Two Chrged Cpcitors Two cpcitors C 1 nd C 2 (where C 1 C 2 ) re chrged to the sme initil potentil difference V i, but with opposite polrity. The chrged cpcitors re removed from the bttery, nd their pltes re connected s shown in Figure The switches S 1 nd S 2 re then closed, s shown in Figure 26.12b. () Find the finl potentil difference V f between nd b fter the switches re closed. Solution Let us identify the left-hnd pltes of the cpcitors s n isolted system becuse they re not connected to the right-hnd pltes by conductors. The chrges on the lefthnd pltes before the switches re closed re Q 1i () C 1 S 1 S 2 Q 2i C 2 b Figure Q C 1 1f S 1 S 2 Q 2f C 2 (b) b Q 1i C 1 V i nd Q 2i C 2 V i The negtive sign for Q 2i is necessry becuse the chrge on the left plte of cpcitor C 2 is negtive. The totl chrge Q in the system is (1) Q Q 1i Q 2i (C 1 C 2 )V i After the switches re closed, the totl chrge in the system remins the sme: (2) Q Q 1f Q 2f The chrges redistribute until the entire system is t the sme potentil V f. Thus, the finl potentil difference cross C 1 must be the sme s the finl potentil difference cross C 2. To stisfy this requirement, the chrges on the cpcitors fter the switches re closed re Q 1f C 1 V f nd Q 2f C 2 V f Dividing the first eqution by the second, we hve Q 1f Q 2f (3) Q 1f C 1 Q 2f C 2 Combining Equtions (2) nd (3), we obtin Q 2 f Q C 2 C 1 C 2 Using Eqution (3) to find Q 1 f in terms of Q, we hve Finlly, using Eqution 26.1 to find the voltge cross ech cpcitor, we find tht V 1f Q 1f C 1 C 1 V f C 2 V f C 1 C 2 Q Q 1f Q 2 f C 1 C 2 Q 2f Q 2f Q 2f 1 C 1 C 2 Q 1f C 1 C 2 Q 2f C 1 C 2 Q C 2 C 1 C 2 Q Q C 1 C 1 C 2 C 1 C 1 C 1 C 2 Q C 1 C 2 As noted erlier, V 1f V 2 f V f. To express V f in terms of the given quntities C 1, C 2, nd V i, we substitute the vlue of Q from Eqution (1) to obtin (b) Find the totl energy stored in the cpcitors before nd fter the switches re closed nd the rtio of the finl energy to the initil energy. Solution Before the switches re closed, the totl energy stored in the cpcitors is U i 1 2 C 1(V i ) C 2(V i ) 2 After the switches re closed, the totl energy stored in the cpcitors is U f 1 2 C 1(V f ) C 2(V f ) (C 1 C 2 )(V f ) 2 Using Eqution (1), we cn express this s U f 1 2 Therefore, the rtio of the finl energy stored to the initil energy stored is U f U i V 2 f Q 2 f C (C 1 C 2 ) Q C 1 C V f C 1 C 2 C 1 C 2 V i Q 2 (C 1 C 2 ) Q C 2 C 1 C 2 (C 1 C 2 ) 2 (V i ) 2 (C 1 C 2 ) C (C 1 C 2 )(V i ) (C 1 C 2 )(V i ) 2 (C 1 C 2 ) 2 (V i ) 2 Q C 1 C 2 Q 2 C 1 C 2 (C 1 C 2 ) C 1 C 2 C 1 C 2 2

113 26.4 Energy Stored in Chrged Cpcitor 817 This rtio is less thn unity, indicting tht the finl energy is less thn the initil energy. At first, you might think tht the lw of energy conservtion hs been violted, but this is not the cse. The missing energy is rdited wy in the form of electromgnetic wves, s we shll see in Chpter 34. Quick Quiz 26.4 You chrge prllel-plte cpcitor, remove it from the bttery, nd prevent the wires connected to the pltes from touching ech other. When you pull the pltes prt, do the following quntities increse, decrese, or sty the sme? () C; (b) Q ; (c) E between the pltes; (d) V ; (e) energy stored in the cpcitor. Quick Quiz 26.5 Repet Quick Quiz 26.4, but this time nswer the questions for the sitution in which the bttery remins connected to the cpcitor while you pull the pltes prt. One device in which cpcitors hve n importnt role is the defibrilltor (Fig ). Up to 360 J is stored in the electric field of lrge cpcitor in defibrilltor when it is fully chrged. The defibrilltor cn deliver ll this energy to ptient in bout 2 ms. (This is roughly equivlent to times the power output of 60-W lightbulb!) The sudden electric shock stops the fibrilltion (rndom contrctions) of the hert tht often ccompnies hert ttcks nd helps to restore the correct rhythm. A cmer s flsh unit lso uses cpcitor, lthough the totl mount of energy stored is much less thn tht stored in defibrilltor. After the flsh unit s cpcitor is chrged, tripping the cmer s shutter cuses the stored energy to be sent through specil lightbulb tht briefly illumintes the subject being photogrphed. web To lern more bout defibrilltors, visit Figure In hospitl or t n emergency scene, you might see ptient being revived with defibrilltor. The defibrilltor s pddles re pplied to the ptient s chest, nd n electric shock is sent through the chest cvity. The im of this technique is to restore the hert s norml rhythm pttern.

114 818 CHAPTER 26 Cpcitnce nd Dielectrics 26.5 CAPACITORS WITH DIELECTRICS A dielectric is nonconducting mteril, such s rubber, glss, or wxed pper. When dielectric is inserted between the pltes of cpcitor, the cpcitnce increses. If the dielectric completely fills the spce between the pltes, the cpcitnce increses by dimensionless fctor, which is clled the dielectric constnt. The dielectric constnt is property of mteril nd vries from one mteril to nother. In this section, we nlyze this chnge in cpcitnce in terms of electricl prmeters such s electric chrge, electric field, nd potentil difference; in Section 26.7, we shll discuss the microscopic origin of these chnges. We cn perform the following experiment to illustrte the effect of dielectric in cpcitor: Consider prllel-plte cpcitor tht without dielectric hs chrge Q 0 nd cpcitnce C 0. The potentil difference cross the cpcitor is V 0 Q 0 /C 0. Figure illustrtes this sitution. The potentil difference is mesured by voltmeter, which we shll study in greter detil in Chpter 28. Note tht no bttery is shown in the figure; lso, we must ssume tht no chrge cn flow through n idel voltmeter, s we shll lern in Section Hence, there is no pth by which chrge cn flow nd lter the chrge on the cpcitor. If dielectric is now inserted between the pltes, s shown in Figure 26.14b, the voltmeter indictes tht the voltge between the pltes decreses to vlue V. The voltges with nd without the dielectric re relted by the fctor s follows: 1. V V 0 Becuse V V 0, we see tht Becuse the chrge Q 0 on the cpcitor does not chnge, we conclude tht the cpcitnce must chnge to the vlue The cpcitnce of filled cpcitor is greter thn tht of n empty one by fctor. C Q 0 V Q 0 V 0 / C C 0 (26.14) Tht is, the cpcitnce increses by the fctor when the dielectric completely fills the region between the pltes. 4 For prllel-plte cpcitor, where C 0 0A/d (Eq. 26.3), we cn express the cpcitnce when the cpcitor is filled with dielectric s 0A C (26.15) d From Equtions 26.3 nd 26.15, it would pper tht we could mke the cpcitnce very lrge by decresing d, the distnce between the pltes. In prctice, the lowest vlue of d is limited by the electric dischrge tht could occur through the dielectric medium seprting the pltes. For ny given seprtion d, the mximum voltge tht cn be pplied to cpcitor without cusing dischrge depends on the dielectric strength (mximum electric field) of the dielectric. If the mgnitude of the electric field in the dielectric exceeds the dielectric strength, then the insulting properties brek down nd the dielectric begins to conduct. Insulting mterils hve vlues of greter thn unity nd dielectric strengths Q 0 V 0 4 If the dielectric is introduced while the potentil difference is being mintined constnt by bttery, the chrge increses to vlue Q Q 0. The dditionl chrge is supplied by the bttery, nd the cpcitnce gin increses by the fctor.

115 26.5 Cpcitors with Dielectrics 819 Dielectric C 0 Q 0 C Q 0 V 0 V Figure () A chrged cpcitor () before nd (b) fter insertion of dielectric between the pltes. The chrge on the pltes remins unchnged, but the potentil difference decreses from V 0 to V V 0 /. Thus, the cpcitnce increses from C 0 to C 0. (b) greter thn tht of ir, s Tble 26.1 indictes. Thus, we see tht dielectric provides the following dvntges: Increse in cpcitnce Increse in mximum operting voltge Possible mechnicl support between the pltes, which llows the pltes to be close together without touching, thereby decresing d nd incresing C TABLE 26.1 Dielectric Constnts nd Dielectric Strengths of Vrious Mterils t Room Temperture Dielectric Dielectric Mteril Constnt Strength (V/m) Air (dry) Bkelite Fused qurtz Neoprene rubber Nylon Pper Polystyrene Polyvinyl chloride Porcelin Pyrex glss Silicone oil Strontium titnte Teflon Vcuum Wter 80 The dielectric strength equls the mximum electric field tht cn exist in dielectric without electricl brekdown. Note tht these vlues depend strongly on the presence of impurities nd flws in the mterils.

116 820 CHAPTER 26 Cpcitnce nd Dielectrics () (b) () Kirlin photogrph creted by dropping steel bll into high-energy electric field. Kirlin photogrphy is lso known s electrophotogrphy. (b) Sprks from sttic electricity dischrge between fork nd four electrodes. Mny sprks were used to crete this imge becuse only one sprk forms for given dischrge. Note tht the bottom prong dischrges to both electrodes t the bottom right. The light of ech sprk is creted by the excittion of gs toms long its pth. Types of Cpcitors Commercil cpcitors re often mde from metllic foil interlced with thin sheets of either prffin-impregnted pper or Mylr s the dielectric mteril. These lternte lyers of metllic foil nd dielectric re rolled into cylinder to form smll pckge (Fig ). High-voltge cpcitors commonly consist of number of interwoven metllic pltes immersed in silicone oil (Fig b). Smll cpcitors re often constructed from cermic mterils. Vrible cpcitors (typiclly 10 to 500 pf) usully consist of two interwoven sets of metllic pltes, one fixed nd the other movble, nd contin ir s the dielectric. Often, n electrolytic cpcitor is used to store lrge mounts of chrge t reltively low voltges. This device, shown in Figure 26.15c, consists of metllic foil in contct with n electrolyte solution tht conducts electricity by virtue of the motion of ions contined in the solution. When voltge is pplied between the foil nd the electrolyte, thin lyer of metl oxide (n insultor) is formed on the foil, Metl foil Pltes Cse Electrolyte Figure Pper () (b) (c) Oil Contcts Metllic foil oxide lyer Three commercil cpcitor designs. () A tubulr cpcitor, whose pltes re seprted by pper nd then rolled into cylinder. (b) A high-voltge cpcitor consisting of mny prllel pltes seprted by insulting oil. (c) An electrolytic cpcitor.

117 26.5 Cpcitors with Dielectrics 821 nd this lyer serves s the dielectric. Very lrge vlues of cpcitnce cn be obtined in n electrolytic cpcitor becuse the dielectric lyer is very thin, nd thus the plte seprtion is very smll. Electrolytic cpcitors re not reversible s re mny other cpcitors they hve polrity, which is indicted by positive nd negtive signs mrked on the device. When electrolytic cpcitors re used in circuits, the polrity must be ligned properly. If the polrity of the pplied voltge is opposite tht which is intended, the oxide lyer is removed nd the cpcitor conducts electricity insted of storing chrge. Quick Quiz 26.6 If you hve ever tried to hng picture, you know it cn be difficult to locte wooden stud in which to nchor your nil or screw. A crpenter s stud-finder is bsiclly cpcitor with its pltes rrnged side by side insted of fcing one nother, s shown in Figure When the device is moved over stud, does the cpcitnce increse or decrese? Stud Cpcitor pltes Figure () Stud-finder Wll bord A stud-finder. ()The mterils between the pltes of the cpcitor re the wllbord nd ir. (b) When the cpcitor moves cross stud in the wll, the mterils between the pltes re the wllbord nd the wood. The chnge in the dielectric constnt cuses signl light to illuminte. (b) EXAMPLE 26.6 A Pper-Filled Cpcitor A prllel-plte cpcitor hs pltes of dimensions 2.0 cm by 3.0 cm seprted by 1.0-mm thickness of pper. () Find its cpcitnce. Solution hve C 0A d F Becuse 3.7 for pper (see Tble 26.1), we (b) Wht is the mximum chrge tht cn be plced on the cpcitor? Solution 3.7( C 2 /Nm 2 ) m 2 20 pf m From Tble 26.1 we see tht the dielectric strength of pper is V/m. Becuse the thickness of the pper is 1.0 mm, the mximum voltge tht cn be pplied before brekdown is Hence, the mximum chrge is Q mx C V mx ( F)( V) Exercise V mx E mx d ( V/m)( m) V Wht is the mximum energy tht cn be stored in the cpcitor? Answer J C

118 822 CHAPTER 26 Cpcitnce nd Dielectrics EXAMPLE 26.7 Energy Stored Before nd After A prllel-plte cpcitor is chrged with bttery to chrge Q 0, s shown in Figure The bttery is then removed, nd slb of mteril tht hs dielectric constnt is inserted between the pltes, s shown in Figure 26.17b. Find the energy stored in the cpcitor before nd fter the dielectric is inserted. Solution The energy stored in the bsence of the dielectric is (see Eq ): U 0 Q 2 0 2C 0 After the bttery is removed nd the dielectric inserted, the chrge on the cpcitor remins the sme. Hence, the energy stored in the presence of the dielectric is U Q 2 0 2C But the cpcitnce in the presence of the dielectric is C C 0, so U becomes U Q 2 0 U 0 2C 0 Becuse 1, the finl energy is less thn the initil energy. We cn ccount for the missing energy by noting tht the dielectric, when inserted, gets pulled into the device (see the following discussion nd Figure 26.18). An externl gent must do negtive work to keep the dielectric from ccelerting. This work is simply the difference U U 0. (Alterntively, the positive work done by the system on the externl gent is U 0 U.) Exercise Suppose tht the cpcitnce in the bsence of dielectric is 8.50 pf nd tht the cpcitor is chrged to potentil difference of 12.0 V. If the bttery is disconnected nd slb of polystyrene is inserted between the pltes, wht is U 0 U? Answer 373 pj. C 0 Q 0 Q 0 () (b) V 0 Dielectric Figure As we hve seen, the energy of cpcitor not connected to bttery is lowered when dielectric is inserted between the pltes; this mens tht negtive work is done on the dielectric by the externl gent inserting the dielectric into the cpcitor. This, in turn, implies tht force tht drws it into the cpcitor must be cting on the dielectric. This force origintes from the nonuniform nture of the electric field of the cpcitor ner its edges, s indicted in Figure The horizontl component of this fringe field cts on the induced chrges on the surfce of the dielectric, producing net horizontl force directed into the spce between the cpcitor pltes. Quick Quiz 26.7 A fully chrged prllel-plte cpcitor remins connected to bttery while you slide dielectric between the pltes. Do the following quntities increse, decrese, or sty the sme? () C; (b) Q ; (c) E between the pltes; (d) V ; (e) energy stored in the cpcitor.

119 26.6 Electric Dipole in n Electric Field 823 Q Q Figure The nonuniform electric field ner the edges of prllel-plte cpcitor cuses dielectric to be pulled into the cpcitor. Note tht the field cts on the induced surfce chrges on the dielectric, which re nonuniformly distributed. Optionl Section 26.6 ELECTRIC DIPOLE IN AN ELECTRIC FIELD We hve discussed the effect on the cpcitnce of plcing dielectric between the pltes of cpcitor. In Section 26.7, we shll describe the microscopic origin of this effect. Before we cn do so, however, we need to expnd upon the discussion of the electric dipole tht we begn in Section 23.4 (see Exmple 23.6). The electric dipole consists of two chrges of equl mgnitude but opposite sign seprted by distnce 2, s shown in Figure The electric dipole moment of this configurtion is defined s the vector p directed from q to q long the line joining the chrges nd hving mgnitude 2q: p 2q (26.16) Now suppose tht n electric dipole is plced in uniform electric field E, s shown in Figure We identify E s the field externl to the dipole, distinguishing it from the field due to the dipole, which we discussed in Section The field E is estblished by some other chrge distribution, nd we plce the dipole into this field. Let us imgine tht the dipole moment mkes n ngle with the field. The electric forces cting on the two chrges re equl in mgnitude but opposite in direction s shown in Figure (ech hs mgnitude F qe). Thus, the net force on the dipole is zero. However, the two forces produce net torque on the dipole; s result, the dipole rottes in the direction tht brings the dipole moment vector into greter lignment with the field. The torque due to the force on the positive chrge bout n xis through O in Figure is F sin, where sin is the moment rm of F bout O. This force tends to produce clockwise rottion. The torque bout O on the negtive chrge lso is F sin ; here gin, the force tends to produce clockwise rottion. Thus, the net torque bout O is 2F sin Becuse F qe nd p 2q, we cn express s 2qE sin pe sin (26.17) q Figure F Figure q O q An electric dipole in uniform externl electric field. The dipole moment p is t n ngle to the field, cusing the dipole to experience torque. p θ q An electric dipole consists of two chrges of equl mgnitude but opposite sign seprted by distnce of 2. The electric dipole moment p is directed from q to q. E F

120 824 CHAPTER 26 Cpcitnce nd Dielectrics It is convenient to express the torque in vector form s the cross product of the vectors p nd E: Torque on n electric dipole in n externl electric field p E (26.18) We cn determine the potentil energy of the system of n electric dipole in n externl electric field s function of the orienttion of the dipole with respect to the field. To do this, we recognize tht work must be done by n externl gent to rotte the dipole through n ngle so s to cuse the dipole moment vector to become less ligned with the field. The work done is then stored s potentil energy in the system of the dipole nd the externl field. The work dw required to rotte the dipole through n ngle d is dw (Eq ). Becuse nd becuse the work is trnsformed into potentil energy U, we find pe sin tht, for rottion from i to f, the chnge in potentil energy is f U f U i i d i f f pe cos pe(cos i cos f) i The term tht contins cos i is constnt tht depends on the initil orienttion of the dipole. It is convenient for us to choose i 90, so tht cos i cos Furthermore, let us choose U i 0 t i 90 s our reference of potentil energy. Hence, we cn express generl vlue of U U f s d p sin d pe i f sin d U pe cos (26.19) We cn write this expression for the potentil energy of dipole in n electric field s the dot product of the vectors p nd E: Potentil energy of dipole in n electric field U p E (26.20) To develop conceptul understnding of Eqution 26.19, let us compre this expression with the expression for the potentil energy of n object in the grvittionl field of the Erth, U mgh (see Chpter 8). The grvittionl expression includes prmeter ssocited with the object we plce in the field its mss m. Likewise, Eqution includes prmeter of the object in the electric field its dipole moment p. The grvittionl expression includes the mgnitude of the grvittionl field g. Similrly, Eqution includes the mgnitude of the electric field E. So fr, these two contributions to the potentil energy expressions pper nlogous. However, the finl contribution is somewht different in the two cses. In the grvittionl expression, the potentil energy depends on how high we lift the object, mesured by h. In Eqution 26.19, the potentil energy depends on the ngle through which we rotte the dipole. In both cses, we re mking chnge in the system. In the grvittionl cse, the chnge involves moving n object in trnsltionl sense, wheres in the electricl cse, the chnge involves moving n object in rottionl sense. In both cses, however, once the chnge is mde, the system tends to return to the originl configurtion when the object is relesed: the object of mss m flls bck to the ground, nd the dipole begins to rotte bck towrd the configurtion in which it ws ligned with the field. Thus, prt from the type of motion, the expressions for potentil energy in these two cses re similr.

121 26.6 Electric Dipole in n Electric Field 825 Molecules re sid to be polrized when seprtion exists between the verge position of the negtive chrges nd the verge position of the positive chrges in the molecule. In some molecules, such s wter, this condition is lwys present such molecules re clled polr molecules. Molecules tht do not possess permnent polriztion re clled nonpolr molecules. We cn understnd the permnent polriztion of wter by inspecting the geometry of the wter molecule. In the wter molecule, the oxygen tom is bonded to the hydrogen toms such tht n ngle of 105 is formed between the two bonds (Fig ). The center of the negtive chrge distribution is ner the oxygen tom, nd the center of the positive chrge distribution lies t point midwy long the line joining the hydrogen toms (the point lbeled in Fig ). We cn model the wter molecule nd other polr molecules s dipoles becuse the verge positions of the positive nd negtive chrges ct s point chrges. As result, we cn pply our discussion of dipoles to the behvior of polr molecules. Microwve ovens tke dvntge of the polr nture of the wter molecule. When in opertion, microwve ovens generte rpidly chnging electric field tht cuses the polr molecules to swing bck nd forth, bsorbing energy from the field in the process. Becuse the jostling molecules collide with ech other, the energy they bsorb from the field is converted to internl energy, which corresponds to n increse in temperture of the food. Another household scenrio in which the dipole structure of wter is exploited is wshing with sop nd wter. Grese nd oil re mde up of nonpolr molecules, which re generlly not ttrcted to wter. Plin wter is not very useful for removing this type of grime. Sop contins long molecules clled surfctnts. In long molecule, the polrity chrcteristics of one end of the molecule cn be different from those t the other end. In surfctnt molecule, one end cts like nonpolr molecule nd the other cts like polr molecule. The nonpolr end cn ttch to grese or oil molecule, nd the polr end cn ttch to wter molecule. Thus, the sop serves s chin, linking the dirt nd wter molecules together. When the wter is rinsed wy, the grese nd oil go with it. A symmetric molecule (Fig ) hs no permnent polriztion, but polriztion cn be induced by plcing the molecule in n electric field. A field directed to the left, s shown in Figure 26.22b, would cuse the center of the positive chrge distribution to shift to the left from its initil position nd the center of the negtive chrge distribution to shift to the right. This induced polriztion is the effect tht predomintes in most mterils used s dielectrics in cpcitors. O H 105 H Figure The wter molecule, H 2 O, hs permnent polriztion resulting from its bent geometry. The center of the positive chrge distribution is t the point. () Figure (b) () A symmetric molecule hs no permnent polriztion. (b) An externl electric field induces polriztion in the molecule. E EXAMPLE 26.8 The H 2 O Molecule The wter (H 2 O) molecule hs n electric dipole moment of C m. A smple contins wter molecules, with the dipole moments ll oriented in the direction of n electric field of mgnitude N/C. How much work is required to rotte the dipoles from this orienttion to one in which ll the dipole moments re perpendiculr to the field ( 0) Solution ( 90)? The work required to rotte one molecule 90 is equl to the difference in potentil energy between the 90 orienttion nd the 0 orienttion. Using Eqution 26.19, we obtin W U 90 U 0 (pe cos 90) (pe cos 0) pe ( Cm)( N/C) J Becuse there re molecules in the smple, the totl work required is W totl (10 21 )( J) J

122 826 CHAPTER 26 Cpcitnce nd Dielectrics Optionl Section 26.7 AN ATOMIC DESCRIPTION OF DIELECTRICS In Section 26.5 we found tht the potentil difference V 0 between the pltes of cpcitor is reduced to V 0 / when dielectric is introduced. Becuse the potentil difference between the pltes equls the product of the electric field nd the seprtion d, the electric field is lso reduced. Thus, if E 0 is the electric field without the dielectric, the field in the presence of dielectric is Figure () (b) E 0 () Polr molecules re rndomly oriented in the bsence of n externl electric field. (b) When n externl field is pplied, the molecules prtilly lign with the field. E E 0 (26.21) Let us first consider dielectric mde up of polr molecules plced in the electric field between the pltes of cpcitor. The dipoles (tht is, the polr molecules mking up the dielectric) re rndomly oriented in the bsence of n electric field, s shown in Figure When n externl field E 0 due to chrges on the cpcitor pltes is pplied, torque is exerted on the dipoles, cusing them to prtilly lign with the field, s shown in Figure 26.23b. We cn now describe the dielectric s being polrized. The degree of lignment of the molecules with the electric field depends on temperture nd on the mgnitude of the field. In generl, the lignment increses with decresing temperture nd with incresing electric field. If the molecules of the dielectric re nonpolr, then the electric field due to the pltes produces some chrge seprtion nd n induced dipole moment. These induced dipole moments tend to lign with the externl field, nd the dielectric is polrized. Thus, we cn polrize dielectric with n externl field regrdless of whether the molecules re polr or nonpolr. With these ides in mind, consider slb of dielectric mteril plced between the pltes of cpcitor so tht it is in uniform electric field E 0, s shown in Figure The electric field due to the pltes is directed to the right nd polrizes the dielectric. The net effect on the dielectric is the formtion of n induced positive surfce chrge density ind on the right fce nd n equl negtive surfce chrge density ind on the left fce, s shown in Figure 26.24b. These induced surfce chrges on the dielectric give rise to n induced electric field E ind in the direction opposite the externl field E 0. Therefore, the net electric field E in the E 0 E 0 E ind σσ ind σ ind Figure () () When dielectric is polrized, the dipole moments of the molecules in the dielectric re prtilly ligned with the externl field E 0. (b) This polriztion cuses n induced negtive surfce chrge on one side of the dielectric nd n equl induced positive surfce chrge on the opposite side. This seprtion of chrge results in reduction in the net electric field within the dielectric. (b)

123 26.7 An Atomic Description of Dielectrics 827 dielectric hs mgnitude E E 0 E ind (26.22) In the prllel-plte cpcitor shown in Figure 26.25, the externl field E 0 is relted to the chrge density on the pltes through the reltionship E 0 /0. The induced electric field in the dielectric is relted to the induced chrge density ind through the reltionship E ind ind/0. Becuse E E 0 / /0, substitution into Eqution gives 1, ind 1 (26.23) Becuse this expression shows tht the chrge density ind induced on the dielectric is less thn the chrge density on the pltes. For instnce, if we see tht the induced chrge density is two-thirds the chrge density on the pltes. If no dielectric is present, then nd ind 0 s expected. However, if the dielectric is replced by n electricl conductor, for which E 0, then Eqution indictes tht E 0 E ind ; this corresponds to ind. Tht is, the surfce chrge induced on the conductor is equl in mgnitude but opposite in sign to tht on the pltes, resulting in net electric field of zero in the conductor. 1 ind 3, σ Figure σσ ind σ ind σ Induced chrge on dielectric plced between the pltes of chrged cpcitor. Note tht the induced chrge density on the dielectric is less thn the chrge density on the pltes. EXAMPLE 26.9 Effect of Metllic Slb A prllel-plte cpcitor hs plte seprtion d nd plte re A. An unchrged metllic slb of thickness is inserted midwy between the pltes. () Find the cpcitnce of the device. Solution We cn solve this problem by noting tht ny chrge tht ppers on one plte of the cpcitor must induce chrge of equl mgnitude but opposite sign on the ner side of the slb, s shown in Figure Consequently, the net chrge on the slb remins zero, nd the electric field inside the slb is zero. Hence, the cpcitor is equivlent to two cpcitors in series, ech hving plte seprtion (d )/2, s shown in Figure 26.26b. Using the rule for dding two cpcitors in series (Eq ), we obtin 1 C 1 C 1 1 C 2 C 0A d 1 0A (d )/2 Note tht C pproches infinity s pproches d. Why? (b) Show tht the cpcitnce is unffected if the metllic slb is infinitesimlly thin. 1 0A (d )/2 Solution In the result for prt (), we let : 0: C lim :0 0A d which is the originl cpcitnce. d Figure σ (d )/2 σ σ (d )/2 σ () 0A d (d )/2 (d )/2 () A prllel-plte cpcitor of plte seprtion d prtilly filled with metllic slb of thickness. (b) The equivlent circuit of the device in prt () consists of two cpcitors in series, ech hving plte seprtion (d )/2. (b)

124 828 CHAPTER 26 Cpcitnce nd Dielectrics (c) Show tht the nswer to prt () does not depend on where the slb is inserted. Solution Let us imgine tht the slb in Figure is moved upwrd so tht the distnce between the upper edge of the slb nd the upper plte is b. Then, the distnce between the lower edge of the slb nd the lower plte is d b. As in prt (), we find the totl cpcitnce of the series combintion: 1 C 1 C 1 1 C 2 1 0A b 0A b d b 0A 0A C d This is the sme result s in prt (). It is independent of the vlue of b, so it does not mtter where the slb is locted. 1 0A d b d 0A EXAMPLE A Prtilly Filled Cpcitor A prllel-plte cpcitor with plte seprtion d hs cpcitnce C 0 in the bsence of dielectric. Wht is the cpcitnce when slb of dielectric mteril of dielectric constnt nd thickness 26.27)? Figure d 3 2 d 3 1 d 3 2 d d is inserted between the pltes (Fig. d () (b) () A prllel-plte cpcitor of plte seprtion d prtilly filled with dielectric of thickness d/3. (b) The equivlent circuit of the cpcitor consists of two cpcitors connected in series. κ κ C 1 C 2 Solution In Exmple 26.9, we found tht we could insert metllic slb between the pltes of cpcitor nd consider the combintion s two cpcitors in series. The resulting cpcitnce ws independent of the loction of the slb. Furthermore, if the thickness of the slb pproches zero, then the cpcitnce of the system pproches the cpcitnce when the slb is bsent. From this, we conclude tht we cn insert n infinitesimlly thin metllic slb nywhere between the pltes of cpcitor without ffecting the cpcitnce. Thus, let us imgine sliding n infinitesimlly thin metllic slb long the bottom fce of the dielectric shown in Figure We cn then consider this system to be the series combintion of the two cpcitors shown in Figure 26.27b: one hving plte seprtion d/3 nd filled with dielectric, nd the other hving plte seprtion 2d/3 nd ir between its pltes. From Equtions nd 26.3, the two cpcitnces re C 1 Using Eqution for two cpcitors combined in series, we hve 1 C 1 1 d/3 C 1 C 2 0A d 30A 1 C 3 0A d/3 nd C A d Becuse the cpcitnce without the dielectric is we see tht C 3 2d/3 0A d 30A 2 1 C 0 0A 2d/3 1 2 C 0 0A/d,

125 Summry 829 SUMMARY A cpcitor consists of two conductors crrying chrges of equl mgnitude but opposite sign. The cpcitnce C of ny cpcitor is the rtio of the chrge Q on either conductor to the potentil difference V between them: C Q V (26.1) This reltionship cn be used in situtions in which ny two of the three vribles re known. It is importnt to remember tht this rtio is constnt for given configurtion of conductors becuse the cpcitnce depends only on the geometry of the conductors nd not on n externl source of chrge or potentil difference. The SI unit of cpcitnce is coulombs per volt, or the frd (F), nd 1 F 1 C/V. Cpcitnce expressions for vrious geometries re summrized in Tble If two or more cpcitors re connected in prllel, then the potentil difference is the sme cross ll of them. The equivlent cpcitnce of prllel combintion of cpcitors is C eq C 1 C 2 C 3 (26.8) If two or more cpcitors re connected in series, the chrge is the sme on ll of them, nd the equivlent cpcitnce of the series combintion is given by (26.10) C eq C 1 C 2 C 3 These two equtions enble you to simplify mny electric circuits by replcing multiple cpcitors with single equivlent cpcitnce. Work is required to chrge cpcitor becuse the chrging process is equivlent to the trnsfer of chrges from one conductor t lower electric potentil to nother conductor t higher potentil. The work done in chrging the cpcitor to chrge Q equls the electric potentil energy U stored in the cpcitor, where U Q 2 2C 1 2 Q V 1 2C(V )2 (26.11) TABLE 26.2 Cpcitnce nd Geometry Geometry Cpcitnce Eqution Isolted chrged sphere of rdius R (second chrged conductor C 40R 26.2 ssumed t infinity) Prllel-plte cpcitor of plte 26.3 re A nd plte seprtion d C A 0 d Cylindricl cpcitor of length C nd inner nd outer rdii 26.4 nd b, respectively 2k e ln b Sphericl cpcitor with inner nd outer rdii nd b, b C 26.6 respectively k e (b )

126 830 CHAPTER 26 Cpcitnce nd Dielectrics When dielectric mteril is inserted between the pltes of cpcitor, the cpcitnce increses by dimensionless fctor, clled the dielectric constnt: C C 0 (26.14) where C 0 is the cpcitnce in the bsence of the dielectric. The increse in cpcitnce is due to decrese in the mgnitude of the electric field in the presence of the dielectric nd to corresponding decrese in the potentil difference between the pltes if we ssume tht the chrging bttery is removed from the circuit before the dielectric is inserted. The decrese in the mgnitude of E rises from n internl electric field produced by ligned dipoles in the dielectric. This internl field produced by the dipoles opposes the pplied field due to the cpcitor pltes, nd the result is reduction in the net electric field. The electric dipole moment p of n electric dipole hs mgnitude p 2q (26.16) The direction of the electric dipole moment vector is from the negtive chrge towrd the positive chrge. The torque cting on n electric dipole in uniform electric field E is (26.18) The potentil energy of n electric dipole in uniform externl electric field E is U pe (26.20) p E Problem-Solving Hints Cpcitors Be creful with units. When you clculte cpcitnce in frds, mke sure tht distnces re expressed in meters nd tht you use the SI vlue of 0. When checking consistency of units, remember tht the unit for electric fields cn be either N/C or V/m. When two or more cpcitors re connected in prllel, the potentil difference cross ech is the sme. The chrge on ech cpcitor is proportionl to its cpcitnce; hence, the cpcitnces cn be dded directly to give the equivlent cpcitnce of the prllel combintion. The equivlent cpcitnce is lwys lrger thn the individul cpcitnces. When two or more cpcitors re connected in series, they crry the sme chrge, nd the sum of the potentil differences equls the totl potentil difference pplied to the combintion. The sum of the reciprocls of the cpcitnces equls the reciprocl of the equivlent cpcitnce, which is lwys less thn the cpcitnce of the smllest individul cpcitor. A dielectric increses the cpcitnce of cpcitor by fctor (the dielectric constnt) over its cpcitnce when ir is between the pltes. For problems in which bttery is being connected or disconnected, note whether modifictions to the cpcitor re mde while it is connected to the bttery or fter it hs been disconnected. If the cpcitor remins connected to the bttery, the voltge cross the cpcitor remins unchnged (equl to the bttery voltge), nd the chrge is proportionl to the cpci-

127 Problems 831 tnce, lthough it my be modified (for instnce, by the insertion of dielectric). If you disconnect the cpcitor from the bttery before mking ny modifictions to the cpcitor, then its chrge remins fixed. In this cse, s you vry the cpcitnce, the voltge cross the pltes chnges ccording to the expression V Q /C. QUESTIONS 1. If you were sked to design cpcitor in sitution for which smll size nd lrge cpcitnce were required, wht fctors would be importnt in your design? 2. The pltes of cpcitor re connected to bttery. Wht hppens to the chrge on the pltes if the connecting wires re removed from the bttery? Wht hppens to the chrge if the wires re removed from the bttery nd connected to ech other? 3. A frd is very lrge unit of cpcitnce. Clculte the length of one side of squre, ir-filled cpcitor tht hs plte seprtion of 1 m. Assume tht it hs cpcitnce of 1 F. 4. A pir of cpcitors re connected in prllel, while n identicl pir re connected in series. Which pir would be more dngerous to hndle fter being connected to the sme voltge source? Explin. 5. If you re given three different cpcitors C 1, C 2, C 3, how mny different combintions of cpcitnce cn you produce? 6. Wht dvntge might there be in using two identicl cpcitors in prllel connected in series with nother identicl prllel pir rther thn single cpcitor? 7. Is it lwys possible to reduce combintion of cpcitors to one equivlent cpcitor with the rules we hve developed? Explin. 8. Becuse the net chrge in cpcitor is lwys zero, wht does cpcitor store? 9. Becuse the chrges on the pltes of prllel-plte cpcitor re of opposite sign, they ttrct ech other. Hence, it would tke positive work to increse the plte seprtion. Wht hppens to the externl work done in this process? 10. Explin why the work needed to move chrge Q through potentil difference V is W Q V, wheres the energy stored in chrged cpcitor is U 1 2 Q V. 1 Where does the 2 fctor come from? 11. If the potentil difference cross cpcitor is doubled, by wht fctor does the stored energy chnge? 12. Why is it dngerous to touch the terminls of highvoltge cpcitor even fter the pplied voltge hs been turned off? Wht cn be done to mke the cpcitor sfe to hndle fter the voltge source hs been removed? 13. Describe how you cn increse the mximum operting voltge of prllel-plte cpcitor for fixed plte seprtion. 14. An ir-filled cpcitor is chrged, disconnected from the power supply, nd, finlly, connected to voltmeter. Explin how nd why the voltge reding chnges when dielectric is inserted between the pltes of the cpcitor. 15. Using the polr molecule description of dielectric, explin how dielectric ffects the electric field inside cpcitor. 16. Explin why dielectric increses the mximum operting voltge of cpcitor even though the physicl size of the cpcitor does not chnge. 17. Wht is the difference between dielectric strength nd the dielectric constnt? 18. Explin why wter molecule is permnently polrized. Wht type of molecule hs no permnent polriztion? 19. If dielectric-filled cpcitor is heted, how does its cpcitnce chnge? (Neglect therml expnsion nd ssume tht the dipole orienttions re temperture dependent.) PROBLEMS 1, 2, 3 = strightforwrd, intermedite, chllenging = full solution vilble in the Student Solutions Mnul nd Study Guide WEB = solution posted t = Computer useful in solving problem = Interctive Physics = pired numericl/symbolic problems Section 26.1 Definition of Cpcitnce 1. () How much chrge is on ech plte of 4.00-F cpcitor when it is connected to 12.0-V bttery? (b) If this sme cpcitor is connected to 1.50-V bttery, wht chrge is stored? 2. Two conductors hving net chrges of 10.0 C nd 10.0 C hve potentil difference of 10.0 V. Determine () the cpcitnce of the system nd (b) the potentil difference between the two conductors if the chrges on ech re incresed to 100 C nd 100 C.

128 832 CHAPTER 26 Cpcitnce nd Dielectrics WEB WEB Section 26.2 Clculting Cpcitnce 3. An isolted chrged conducting sphere of rdius 12.0 cm cretes n electric field of N/C t distnce 21.0 cm from its center. () Wht is its surfce chrge density? (b) Wht is its cpcitnce? 4. () If drop of liquid hs cpcitnce 1.00 pf, wht is its rdius? (b) If nother drop hs rdius 2.00 mm, wht is its cpcitnce? (c) Wht is the chrge on the smller drop if its potentil is 100 V? 5. Two conducting spheres with dimeters of m nd 1.00 m re seprted by distnce tht is lrge compred with the dimeters. The spheres re connected by thin wire nd re chrged to 7.00 C. () How is this totl chrge shred between the spheres? (Neglect ny chrge on the wire.) (b) Wht is the potentil of the system of spheres when the reference potentil is tken to be V 0 t r? 6. Regrding the Erth nd cloud lyer 800 m bove the Erth s the pltes of cpcitor, clculte the cpcitnce if the cloud lyer hs n re of 1.00 km 2. Assume tht the ir between the cloud nd the ground is pure nd dry. Assume tht chrge builds up on the cloud nd on the ground until uniform electric field with mgnitude of N/C throughout the spce between them mkes the ir brek down nd conduct electricity s lightning bolt. Wht is the mximum chrge the cloud cn hold? 7. An ir-filled cpcitor consists of two prllel pltes, ech with n re of 7.60 cm 2, seprted by distnce of 1.80 mm. If 20.0-V potentil difference is pplied to these pltes, clculte () the electric field between the pltes, (b) the surfce chrge density, (c) the cpcitnce, nd (d) the chrge on ech plte. 8. A 1-megbit computer memory chip contins mny 60.0-fF cpcitors. Ech cpcitor hs plte re of m 2. Determine the plte seprtion of such cpcitor (ssume prllel-plte configurtion). The chrcteristic tomic dimeter is m nm. Express the plte seprtion in nnometers. 9. When potentil difference of 150 V is pplied to the pltes of prllel-plte cpcitor, the pltes crry surfce chrge density of 30.0 nc/cm 2. Wht is the spcing between the pltes? 10. A vrible ir cpcitor used in tuning circuits is mde of N semicirculr pltes ech of rdius R nd positioned distnce d from ech other. As shown in Figure P26.10, second identicl set of pltes is enmeshed with its pltes hlfwy between those of the first set. The second set cn rotte s unit. Determine the cpcitnce s function of the ngle of rottion, where corresponds to the mximum cpcitnce. 11. A 50.0-m length of coxil cble hs n inner conductor tht hs dimeter of 2.58 mm nd crries chrge of 8.10 C. The surrounding conductor hs n inner dimeter of 7.27 mm nd chrge of 8.10 C. () Wht is the cpcitnce of this cble? (b) Wht is 0 d the potentil difference between the two conductors? Assume the region between the conductors is ir. 12. A 20.0-F sphericl cpcitor is composed of two metllic spheres, one hving rdius twice s lrge s the other. If the region between the spheres is vcuum, determine the volume of this region. 13. A smll object with mss of 350 mg crries chrge of 30.0 nc nd is suspended by thred between the verticl pltes of prllel-plte cpcitor. The pltes re seprted by 4.00 cm. If the thred mkes n ngle of 15.0 with the verticl, wht is the potentil difference between the pltes? 14. A smll object of mss m crries chrge q nd is suspended by thred between the verticl pltes of prllel-plte cpcitor. The plte seprtion is d. If the thred mkes n ngle with the verticl, wht is the potentil difference between the pltes? 15. An ir-filled sphericl cpcitor is constructed with inner nd outer shell rdii of 7.00 nd 14.0 cm, respectively. () Clculte the cpcitnce of the device. (b) Wht potentil difference between the spheres results in chrge of 4.00 C on the cpcitor? 16. Find the cpcitnce of the Erth. (Hint: The outer conductor of the sphericl cpcitor my be considered s conducting sphere t infinity where V pproches zero.) Section 26.3 Combintions of Cpcitors 17. Two cpcitors C F nd C F re connected in prllel, nd the resulting combintion is connected to 9.00-V bttery. () Wht is the vlue of the equivlent cpcitnce of the combintion? Wht re (b) the potentil difference cross ech cpcitor nd (c) the chrge stored on ech cpcitor? 18. The two cpcitors of Problem 17 re now connected in series nd to 9.00-V bttery. Find () the vlue of the equivlent cpcitnce of the combintion, (b) the voltge cross ech cpcitor, nd (c) the chrge on ech cpcitor. 19. Two cpcitors when connected in prllel give n equivlent cpcitnce of 9.00 pf nd n equivlent c- R Figure P26.10

129 Problems 833 WEB pcitnce of 2.00 pf when connected in series. Wht is the cpcitnce of ech cpcitor? 20. Two cpcitors when connected in prllel give n equivlent cpcitnce of C p nd n equivlent cpcitnce of C s when connected in series. Wht is the cpcitnce of ech cpcitor? 21. Four cpcitors re connected s shown in Figure P () Find the equivlent cpcitnce between points nd b. (b) Clculte the chrge on ech cpcitor if V b 15.0 V µf µ 3.00 µf µ 6.00 µf µ Figure P µf µ b 24. According to its design specifiction, the timer circuit delying the closing of n elevtor door is to hve cpcitnce of 32.0 F between two points A nd B. () When one circuit is being constructed, the inexpensive cpcitor instlled between these two points is found to hve cpcitnce 34.8 F. To meet the specifiction, one dditionl cpcitor cn be plced between the two points. Should it be in series or in prllel with the 34.8-F cpcitor? Wht should be its cpcitnce? (b) The next circuit comes down the ssembly line with cpcitnce 29.8 F between A nd B. Wht dditionl cpcitor should be instlled in series or in prllel in tht circuit, to meet the specifiction? 25. The circuit in Figure P26.25 consists of two identicl prllel metllic pltes connected by identicl metllic springs to 100-V bttery. With the switch open, the pltes re unchrged, re seprted by distnce d 8.00 mm, nd hve cpcitnce C 2.00 F. When the switch is closed, the distnce between the pltes decreses by fctor of () How much chrge collects on ech plte nd (b) wht is the spring constnt for ech spring? (Hint: Use the result of Problem 35.) 22. Evlute the equivlent cpcitnce of the configurtion shown in Figure P All the cpcitors re identicl, nd ech hs cpcitnce C. k d k C C C C C C S 23. Consider the circuit shown in Figure P26.23, where C F, C F, nd V 20.0 V. Cpcitor C 1 is first chrged by the closing of switch S 1. Switch S 1 is then opened, nd the chrged cpcitor is connected to the unchrged cpcitor by the closing of S 2. Clculte the initil chrge cquired by C 1 nd the finl chrge on ech. V Figure P26.22 S 1 C 1 C 2 S 2 Figure P Figure P26.26 shows six concentric conducting spheres, A, B, C, D, E, nd F hving rdii R, 2R, 3R, 4R, 5R, nd 6R, respectively. Spheres B nd C re connected by conducting wire, s re spheres D nd E. Determine the equivlent cpcitnce of this system. 27. A group of identicl cpcitors is connected first in series nd then in prllel. The combined cpcitnce in prllel is 100 times lrger thn for the series connection. How mny cpcitors re in the group? 28. Find the equivlent cpcitnce between points nd b for the group of cpcitors connected s shown in Figure P26.28 if C F, C F, nd C F. 29. For the network described in the previous problem if the potentil difference between points nd b is 60.0 V, wht chrge is stored on C 3? V Figure P26.25

130 834 CHAPTER 26 Cpcitnce nd Dielectrics A B C D E F Figure P26.26 C 1 C 1 C 2 C 3 C 2 WEB energy stored in the two cpcitors. (b) Wht potentil difference would be required cross the sme two cpcitors connected in series so tht the combintion stores the sme energy s in prt ()? Drw circuit digrm of this circuit. 33. A prllel-plte cpcitor is chrged nd then disconnected from bttery. By wht frction does the stored energy chnge (increse or decrese) when the plte seprtion is doubled? 34. A uniform electric field E V/m exists within certin region. Wht volume of spce contins n energy equl to J? Express your nswer in cubic meters nd in liters. 35. A prllel-plte cpcitor hs chrge Q nd pltes of re A. Show tht the force exerted on ech plte by the other is F Q 2 /20A. (Hint: Let C 0A/x for n rbitrry plte seprtion x ; then require tht the work done in seprting the two chrged pltes be W F dx.) 36. Plte of prllel-plte, ir-filled cpcitor is connected to spring hving force constnt k, nd plte b is fixed. They rest on tble top s shown (top view) in Figure P If chrge Q is plced on plte nd chrge Q is plced on plte b, by how much does the spring expnd? C 2 C 2 Figure P26.28 Problems 28 nd Find the equivlent cpcitnce between points nd b in the combintion of cpcitors shown in Figure P µfµ 7.0 µfµ 5.0 µf µ 6.0 µf µ Section 26.4 Energy Stored in Chrged Cpcitor 31. () A 3.00-F cpcitor is connected to 12.0-V bttery. How much energy is stored in the cpcitor? (b) If the cpcitor hd been connected to 6.00-V bttery, how much energy would hve been stored? 32. Two cpcitors C F nd C F re connected in prllel nd chrged with 100-V power supply. () Drw circuit digrm nd clculte the totl b Figure P26.30 b k 37. Review Problem. A certin storm cloud hs potentil difference of V reltive to tree. If, during lightning storm, 50.0 C of chrge is trnsferred through this potentil difference nd 1.00% of the energy is bsorbed by the tree, how much wter (sp in the tree) initilly t 30.0 C cn be boiled wy? Wter hs specific het of J/kg C, boiling point of 100 C, nd het of vporiztion of J/kg. 38. Show tht the energy ssocited with conducting sphere of rdius R nd chrge Q surrounded by vcuum is U k e Q 2 /2R. 39. Einstein sid tht energy is ssocited with mss ccording to the fmous reltionship E mc 2. Estimte the rdius of n electron, ssuming tht its chrge is distributed uniformly over the surfce of sphere of rdius R nd tht the mssenergy of the electron is equl to the totl energy stored in the resulting nonzero electric field between R nd infinity. (See Problem 38. Experimentlly, n electron nevertheless ppers to be point prticle. The electric field close to the electron must be described by quntum electrodynmics, rther thn the clssicl electrodynmics tht we study.) b Figure P26.36

131 Problems 835 Section 26.5 Cpcitors with Dielectrics 40. Find the cpcitnce of prllel-plte cpcitor tht uses Bkelite s dielectric, if ech of the pltes hs n re of 5.00 cm 2 nd the plte seprtion is 2.00 mm. 41. Determine () the cpcitnce nd (b) the mximum voltge tht cn be pplied to Teflon-filled prllelplte cpcitor hving plte re of 1.75 cm 2 nd plte seprtion of mm. 42. () How much chrge cn be plced on cpcitor with ir between the pltes before it breks down, if the re of ech of the pltes is 5.00 cm 2? (b) Find the mximum chrge if polystyrene is used between the pltes insted of ir. 43. A commercil cpcitor is constructed s shown in Figure This prticulr cpcitor is rolled from two strips of luminum seprted by two strips of prffincoted pper. Ech strip of foil nd pper is 7.00 cm wide. The foil is mm thick, nd the pper is mm thick nd hs dielectric constnt of Wht length should the strips be if cpcitnce of F is desired? (Use the prllel-plte formul.) 44. The supermrket sells rolls of luminum foil, plstic wrp, nd wxed pper. Describe cpcitor mde from supermrket mterils. Compute order-of-mgnitude estimtes for its cpcitnce nd its brekdown voltge. 45. A cpcitor tht hs ir between its pltes is connected cross potentil difference of 12.0 V nd stores 48.0 C of chrge. It is then disconnected from the source while still chrged. () Find the cpcitnce of the cpcitor. (b) A piece of Teflon is inserted between the pltes. Find its new cpcitnce. (c) Find the voltge nd chrge now on the cpcitor. 46. A prllel-plte cpcitor in ir hs plte seprtion of 1.50 cm nd plte re of 25.0 cm 2. The pltes re chrged to potentil difference of 250 V nd disconnected from the source. The cpcitor is then immersed in distilled wter. Determine () the chrge on the pltes before nd fter immersion, (b) the cpcitnce nd voltge fter immersion, nd (c) the chnge in energy of the cpcitor. Neglect the conductnce of the liquid. 47. A conducting sphericl shell hs inner rdius nd outer rdius c. The spce between these two surfces is filled with dielectric for which the dielectric constnt is 1 between nd b, nd 2 between b nd c (Fig. P26.47). Determine the cpcitnce of this system. 48. A wfer of titnium dioxide hs n re of 1.00 cm 2 nd thickness of mm. Aluminum is evported on the prllel fces to form prllel-plte cpcitor. () Clculte the cpcitnce. (b) When the cpcitor is chrged with 12.0-V bttery, wht is the mgnitude of chrge delivered to ech plte? (c) For the sitution in prt (b), wht re the free nd induced surfce chrge densities? (d) Wht is the mgnitude E of the electric field? ( 173) 49. Ech cpcitor in the combintion shown in Figure P26.49 hs brekdown voltge of 15.0 V. Wht is the brekdown voltge of the combintion? 20.0 µf µ 20.0 µf µ Q κ 2 κ µf µ 20.0 µf µ 20.0 µf µ (Optionl) Section 26.6 Electric Dipole in n Electric Field 50. A smll rigid object crries positive nd negtive 3.50-nC chrges. It is oriented so tht the positive chrge is t the point ( 1.20 mm, 1.10 mm) nd the negtive chrge is t the point (1.40 mm, 1.30 mm). () Find the electric dipole moment of the object. The object is plced in n electric field E (7 800i 4 900j) N/C. (b) Find the torque cting on the object. (c) Find the potentil energy of the object in this orienttion. (d) If the orienttion of the object cn chnge, find the difference between its mximum nd its minimum potentil energies. 51. A smll object with electric dipole moment p is plced in nonuniform electric field E E(x)i. Tht is, the field is in the x direction, nd its mgnitude depends on the coordinte x. Let represent the ngle between the dipole moment nd the x direction. () Prove tht the dipole experiences net force F p(de/dx) cos in the direction towrd which the field increses. (b) Consider the field creted by sphericl blloon centered t the origin. The blloon hs rdius of 15.0 cm nd crries chrge of 2.00 C. Evlute de/dx t the point (16 cm, 0, 0). Assume tht wter droplet t this point hs n induced dipole moment of (6.30i) nc m. Find the force on it. (Optionl) Section 26.7 An Atomic Description of Dielectrics 52. A detector of rdition clled GeigerMuller counter consists of closed, hollow, conducting cylinder with Q c b Figure P26.47 Figure P26.49

132 836 CHAPTER 26 Cpcitnce nd Dielectrics fine wire long its xis. Suppose tht the internl dimeter of the cylinder is 2.50 cm nd tht the wire long the xis hs dimeter of mm. If the dielectric strength of the gs between the centrl wire nd the cylinder is V/m, clculte the mximum voltge tht cn be pplied between the wire nd the cylinder before brekdown occurs in the gs. 53. The generl form of Guss s lw describes how chrge cretes n electric field in mteril, s well s in vcuum. It is E da q where 0 is the permittivity of the mteril. () A sheet with chrge Q uniformly distributed over its re A is surrounded by dielectric. Show tht the sheet cretes uniform electric field with mgnitude E Q /2A t nerby points. (b) Two lrge sheets of re A crrying opposite chrges of equl mgnitude Q re smll distnce d prt. Show tht they crete uniform electric field of mgnitude E Q /A between them. (c) Assume tht the negtive plte is t zero potentil. Show tht the positive plte is t potentil Qd /A. (d) Show tht the cpcitnce of the pir of pltes is A/d A0/d. ADDITIONAL PROBLEMS 54. For the system of cpcitors shown in Figure P26.54, find () the equivlent cpcitnce of the system, (b) the potentil difference cross ech cpcitor, (c) the chrge on ech cpcitor, nd (d) the totl energy stored by the group µf µ 6.00 µf µ 2.00 µf µ 4.00 µf µ WEB 56. A 2.00-nF prllel-plte cpcitor is chrged to n initil potentil difference V i 100 V nd then isolted. The dielectric mteril between the pltes is mic ( 5.00). () How much work is required to withdrw the mic sheet? (b) Wht is the potentil difference of the cpcitor fter the mic is withdrwn? 57. A prllel-plte cpcitor is constructed using dielectric mteril whose dielectric constnt is 3.00 nd whose dielectric strength is V/m. The desired cpcitnce is F, nd the cpcitor must withstnd mximum potentil difference of V. Find the minimum re of the cpcitor pltes. 58. A prllel-plte cpcitor is constructed using three dielectric mterils, s shown in Figure P You my ssume tht W d. () Find n expression for the cpcitnce of the device in terms of the plte re A nd d, 1, 2, nd 3. (b) Clculte the cpcitnce using the vlues A 1.00 cm 2, d 2.00 mm, , , nd d κ 1 /2 59. A conducting slb of thickness d nd re A is inserted into the spce between the pltes of prllel-plte cpcitor with spcing s nd surfce re A, s shown in Figure P The slb is not necessrily hlfwy between the cpcitor pltes. Wht is the cpcitnce of the system? κ 2 κ 3 Figure P26.58 A d/ V Figure P26.54 s A Figure P26.59 d 55. Consider two long, prllel, nd oppositely chrged wires of rdius d with their centers seprted by distnce D. Assuming the chrge is distributed uniformly on the surfce of ech wire, show tht the cpcitnce per unit length of this pir of wires is C 0 ln D d d 60. () Two spheres hve rdii nd b nd their centers re distnce d prt. Show tht the cpcitnce of this system is C b 2 d provided tht d is lrge compred with nd b. (Hint: Becuse the spheres re fr prt, ssume tht the

133 Problems 837 chrge on one sphere does not perturb the chrge distribution on the other sphere. Thus, the potentil of ech sphere is expressed s tht of symmetric chrge distribution, V k e Q /r, nd the totl potentil t ech sphere is the sum of the potentils due to ech sphere. (b) Show tht s d pproches infinity the bove result reduces to tht of two isolted spheres in series. 61. When certin ir-filled prllel-plte cpcitor is connected cross bttery, it cquires chrge (on ech plte) of q 0. While the bttery connection is mintined, dielectric slb is inserted into nd fills the region between the pltes. This results in the ccumultion of n dditionl chrge q on ech plte. Wht is the dielectric constnt of the slb? 62. A cpcitor is constructed from two squre pltes of sides nd seprtion d. A mteril of dielectric constnt is inserted distnce x into the cpcitor, s shown in Figure P () Find the equivlent cpcitnce of the device. (b) Clculte the energy stored in the cpcitor if the potentil difference is V. (c) Find the direction nd mgnitude of the force exerted on the dielectric, ssuming constnt potentil difference V. Neglect friction. (d) Obtin numericl vlue for the force ssuming tht 5.00 cm, V V, d 2.00 mm, nd the dielectric is glss ( 4.50). (Hint: The system cn be considered s two cpcitors connected in prllel.) 64. When considering the energy supply for n utomobile, the energy per unit mss of the energy source is n importnt prmeter. Using the following dt, compre the energy per unit mss ( J/kg) for gsoline, ledcid btteries, nd cpcitors. (The mpere A will be introduced in Chpter 27 nd is the SI unit of electric current. 1 A 1 C/s.) Gsoline: Btu/gl; density 670 kg/m 3 Ledcid bttery: 12.0 V; 100 A h; mss 16.0 kg Cpcitor: potentil difference t full chrge 12.0 V; cpcitnce F; mss kg 65. An isolted cpcitor of unknown cpcitnce hs been chrged to potentil difference of 100 V. When the chrged cpcitor is then connected in prllel to n unchrged 10.0-F cpcitor, the voltge cross the combintion is 30.0 V. Clculte the unknown cpcitnce. 66. A certin electronic circuit clls for cpcitor hving cpcitnce of 1.20 pf nd brekdown potentil of V. If you hve supply of 6.00-pF cpcitors, ech hving brekdown potentil of 200 V, how could you meet this circuit requirement? 67. In the rrngement shown in Figure P26.67, potentil difference V is pplied, nd C 1 is djusted so tht the voltmeter between points b nd d reds zero. This blnce occurs when C F. If C F nd C F, clculte the vlue of C 2. κ x d C 4 C 1 Figure P26.62 Problems 62 nd 63. V d V b 63. A cpcitor is constructed from two squre pltes of sides nd seprtion d, s suggested in Figure P You my ssume tht d is much less thn. The pltes crry chrges Q 0 nd Q 0. A block of metl hs width, length, nd thickness slightly less thn d. It is inserted distnce x into the cpcitor. The chrges on the pltes re not disturbed s the block slides in. In sttic sitution, metl prevents n electric field from penetrting it. The metl cn be thought of s perfect dielectric, with :. () Clculte the stored energy s function of x. (b) Find the direction nd mgnitude of the force tht cts on the metllic block. (c) The re of the dvncing front fce of the block is essentilly equl to d. Considering the force on the block s cting on this fce, find the stress (force per re) on it. (d) For comprison, express the energy density in the electric field between the cpcitor pltes in terms of Q 0,, d, nd 0. C 3 Figure P It is possible to obtin lrge potentil differences by first chrging group of cpcitors connected in prllel nd then ctivting switch rrngement tht in effect disconnects the cpcitors from the chrging source nd from ech other nd reconnects them in series rrngement. The group of chrged cpcitors is then dischrged in series. Wht is the mximum potentil difference tht cn be obtined in this mnner by using ten cpcitors ech of 500 F nd chrging source of 800 V? 69. A prllel-plte cpcitor of plte seprtion d is chrged to potentil difference V 0. A dielectric slb c C 2

134 838 CHAPTER 26 Cpcitnce nd Dielectrics of thickness d nd dielectric constnt is introduced between the pltes while the bttery remins connected to the pltes. () Show tht the rtio of energy stored fter the dielectric is introduced to the energy stored in the empty cpcitor is U/U 0. Give physicl explntion for this increse in stored energy. (b) Wht hppens to the chrge on the cpcitor? (Note tht this sitution is not the sme s Exmple 26.7, in which the bttery ws removed from the circuit before the dielectric ws introduced.) 70. A prllel-plte cpcitor with pltes of re A nd plte seprtion d hs the region between the pltes filled with two dielectric mterils s in Figure P Assume tht d V L nd tht d V W. () Determine the cpcitnce nd (b) show tht when 1 2 your result becomes the sme s tht for cpcitor contining single dielectric, C 0A/d. d κ 1 L W pcitors re disconnected from the bttery nd from ech other. They re then connected positive plte to negtive plte nd negtive plte to positive plte. Clculte the resulting chrge on ech cpcitor. 73. The inner conductor of coxil cble hs rdius of mm, nd the outer conductor s inside rdius is 3.00 mm. The spce between the conductors is filled with polyethylene, which hs dielectric constnt of 2.30 nd dielectric strength of V/m. Wht is the mximum potentil difference tht this cble cn withstnd? 74. You re optimizing coxil cble design for mjor mnufcturer. Show tht for given outer conductor rdius b, mximum potentil difference cpbility is ttined when the rdius of the inner conductor is b/e where e is the bse of nturl logrithms. 75. Clculte the equivlent cpcitnce between the points nd b in Figure P Note tht this is not simple series or prllel combintion. (Hint: Assume potentil difference V between points nd b. Write expressions for V b in terms of the chrges nd cpcitnces for the vrious possible pthwys from to b, nd require conservtion of chrge for those cpcitor pltes tht re connected to ech other.) κ 2 Figure P µf µ 71. A verticl prllel-plte cpcitor is hlf filled with dielectric for which the dielectric constnt is 2.00 (Fig. P26.71). When this cpcitor is positioned horizontlly, wht frction of it should be filled with the sme dielectric (Fig. P26.71b) so tht the two cpcitors hve equl cpcitnce? 2.00 µf µ 8.00 µf µ 4.00 µfµ 2.00 µf µ b Figure P Determine the effective cpcitnce of the combintion shown in Figure P (Hint: Consider the symmetry involved!) C 2C () Figure P26.71 (b) 3C 72. Cpcitors C F nd C F re chrged s prllel combintion cross 250-V bttery. The c- C 2C Figure P26.76

135 Answers to Quick Quizzes 839 ANSWERS TO QUICK QUIZZES 26.1 () becuse the plte seprtion is decresed. Cpcitnce depends only on how cpcitor is constructed nd not on the externl circuit Zero. If you construct sphericl gussin surfce outside nd concentric with the cpcitor, the net chrge inside the surfce is zero. Applying Guss s lw to this configurtion, we find tht E 0 t points outside the cpcitor For given voltge, the energy stored in cpcitor is proportionl to C: U C(V ) 2 /2. Thus, you wnt to mximize the equivlent cpcitnce. You do this by connecting the three cpcitors in prllel, so tht the cpcitnces dd () C decreses (Eq. 26.3). (b) Q stys the sme becuse there is no plce for the chrge to flow. (c) E remins constnt (see Eq nd the prgrph following it). (d) V increses becuse V Q /C, Q is constnt (prt b), nd C decreses (prt ). (e) The energy stored in the cpcitor is proportionl to both Q nd V (Eq ) nd thus increses. The dditionl energy comes from the work you do in pulling the two pltes prt () C decreses (Eq. 26.3). (b) Q decreses. The bttery supplies constnt potentil difference V ; thus, chrge must flow out of the cpcitor if C Q /V is to de- crese. (c) E decreses becuse the chrge density on the pltes decreses. (d) V remins constnt becuse of the presence of the bttery. (e) The energy stored in the cpcitor decreses (Eq ) It increses. The dielectric constnt of wood (nd of ll other insulting mterils, for tht mtter) is greter thn 1; therefore, the cpcitnce increses (Eq ). This increse is sensed by the stud-finder s specil circuitry, which cuses n indictor on the device to light up () C increses (Eq ). (b) Q increses. Becuse the bttery mintins constnt V, Q must increse if C (Q /V ) increses. (c) E between the pltes remins constnt becuse V Ed nd neither V nor d chnges. The electric field due to the chrges on the pltes increses becuse more chrge hs flowed onto the pltes. The induced surfce chrges on the dielectric crete field tht opposes the increse in the field cused by the greter number of chrges on the pltes. (d) The bttery mintins constnt V. (e) The energy stored in the cpcitor increses (Eq ). You would hve to push the dielectric into the cpcitor, just s you would hve to do positive work to rise mss nd increse its grvittionl potentil energy.

136

137 P U Z Z L E R Electricl workers restoring power to the estern Ontrio town of St. Isdore, which ws without power for severl dys in Jnury 1998 becuse of severe ice storm. It is very dngerous to touch fllen power trnsmission lines becuse of their high electric potentil, which might be hundreds of thousnds of volts reltive to the ground. Why is such high potentil difference used in power trnsmission if it is so dngerous, nd why ren t birds tht perch on the wires electrocuted? (AP/Wide World Photos/Fred Chrtrnd) c h p t e r Current nd Resistnce Chpter Outline 27.1 Electric Current 27.2 Resistnce nd Ohm s Lw 27.3 A Model for Electricl Conduction 27.4 Resistnce nd Temperture 27.5 (Optionl) Superconductors 27.6 Electricl Energy nd Power 840

138 27.1 Electric Current 841 T hus fr our tretment of electricl phenomen hs been confined to the study of chrges t rest, or electrosttics. We now consider situtions involving electric chrges in motion. We use the term electric current, or simply current, to describe the rte of flow of chrge through some region of spce. Most prcticl pplictions of electricity del with electric currents. For exmple, the bttery in flshlight supplies current to the filment of the bulb when the switch is turned on. A vriety of home pplinces operte on lternting current. In these common situtions, the chrges flow through conductor, such s copper wire. It lso is possible for currents to exist outside conductor. For instnce, bem of electrons in television picture tube constitutes current. This chpter begins with the definitions of current nd current density. A microscopic description of current is given, nd some of the fctors tht contribute to the resistnce to the flow of chrge in conductors re discussed. A clssicl model is used to describe electricl conduction in metls, nd some of the limittions of this model re cited ELECTRIC CURRENT It is instructive to drw n nlogy between wter flow nd current. In mny loclities it is common prctice to instll low-flow showerheds in homes s wterconservtion mesure. We quntify the flow of wter from these nd similr devices by specifying the mount of wter tht emerges during given time intervl, which is often mesured in liters per minute. On grnder scle, we cn chrcterize river current by describing the rte t which the wter flows pst prticulr loction. For exmple, the flow over the brink t Nigr Flls is mintined t rtes between m 3 /s nd m 3 /s. Now consider system of electric chrges in motion. Whenever there is net flow of chrge through some region, current is sid to exist. To define current more precisely, suppose tht the chrges re moving perpendiculr to surfce of re A, s shown in Figure (This re could be the cross-sectionl re of wire, for exmple.) The current is the rte t which chrge flows through this surfce. If Q is the mount of chrge tht psses through this re in time intervl t, the verge current I v is equl to the chrge tht psses through A per unit time: I v Q (27.1) t If the rte t which chrge flows vries in time, then the current vries in time; we define the instntneous current I s the differentil limit of verge current: Figure 27.1 A Chrges in motion through n re A. The time rte t which chrge flows through the re is defined s the current I. The direction of the current is the direction in which positive chrges flow when free to do so. I I dq dt (27.2) Electric current The SI unit of current is the mpere (A): 1 A 1 C (27.3) 1 s Tht is, 1 A of current is equivlent to 1 C of chrge pssing through the surfce re in 1 s. The chrges pssing through the surfce in Figure 27.1 cn be positive or negtive, or both. It is conventionl to ssign to the current the sme direction s the flow of positive chrge. In electricl conductors, such s copper or lu- The direction of the current

139 842 CHAPTER 27 Current nd Resistnce minum, the current is due to the motion of negtively chrged electrons. Therefore, when we spek of current in n ordinry conductor, the direction of the current is opposite the direction of flow of electrons. However, if we re considering bem of positively chrged protons in n ccelertor, the current is in the direction of motion of the protons. In some cses such s those involving gses nd electrolytes, for instnce the current is the result of the flow of both positive nd negtive chrges. If the ends of conducting wire re connected to form loop, ll points on the loop re t the sme electric potentil, nd hence the electric field is zero within nd t the surfce of the conductor. Becuse the electric field is zero, there is no net trnsport of chrge through the wire, nd therefore there is no current. The current in the conductor is zero even if the conductor hs n excess of chrge on it. However, if the ends of the conducting wire re connected to bttery, ll points on the loop re not t the sme potentil. The bttery sets up potentil difference between the ends of the loop, creting n electric field within the wire. The electric field exerts forces on the conduction electrons in the wire, cusing them to move round the loop nd thus creting current. It is common to refer to moving chrge (positive or negtive) s mobile chrge crrier. For exmple, the mobile chrge crriers in metl re electrons. Figure 27.2 q x v d t A section of uniform conductor of cross-sectionl re A. The mobile chrge crriers move with speed v d, nd the distnce they trvel in time t is x v d t. The number of crriers in the section of length x is nav d t, where n is the number of crriers per unit volume. v d Averge current in conductor A Microscopic Model of Current We cn relte current to the motion of the chrge crriers by describing microscopic model of conduction in metl. Consider the current in conductor of cross-sectionl re A (Fig. 27.2). The volume of section of the conductor of length x (the gry region shown in Fig. 27.2) is A x. If n represents the number of mobile chrge crriers per unit volume (in other words, the chrge crrier density), the number of crriers in the gry section is na x. Therefore, the chrge Q in this section is Q number of crriers in section chrge per crrier (na x)q where q is the chrge on ech crrier. If the crriers move with speed v d, the distnce they move in time t is x v d t. Therefore, we cn write Q in the form Q (nav d t)q If we divide both sides of this eqution by t, we see tht the verge current in the conductor is I v Q nqv d A (27.4) t The speed of the chrge crriers v d is n verge speed clled the drift speed. To understnd the mening of drift speed, consider conductor in which the chrge crriers re free electrons. If the conductor is isolted tht is, the potentil difference cross it is zero then these electrons undergo rndom motion tht is nlogous to the motion of gs molecules. As we discussed erlier, when potentil difference is pplied cross the conductor (for exmple, by mens of bttery), n electric field is set up in the conductor; this field exerts n electric force on the electrons, producing current. However, the electrons do not move in stright lines long the conductor. Insted, they collide repetedly with the metl toms, nd their resultnt motion is complicted nd zigzg (Fig. 27.3). Despite the collisions, the electrons move slowly long the conductor (in direction opposite tht of E) t the drift velocity v d.

140 27.1 Electric Current 843 v d E Figure 27.3 A schemtic representtion of the zigzg motion of n electron in conductor. The chnges in direction re the result of collisions between the electron nd toms in the conductor. Note tht the net motion of the electron is opposite the direction of the electric field. Ech section of the zigzg pth is prbolic segment. We cn think of the tomelectron collisions in conductor s n effective internl friction (or drg force) similr to tht experienced by the molecules of liquid flowing through pipe stuffed with steel wool. The energy trnsferred from the electrons to the metl toms during collision cuses n increse in the vibrtionl energy of the toms nd corresponding increse in the temperture of the conductor. Quick Quiz 27.1 Consider positive nd negtive chrges moving horizontlly through the four regions shown in Figure Rnk the current in these four regions, from lowest to highest. () (b) (c) (d) Figure 27.4 EXAMPLE 27.1 Drift Speed in Copper Wire The 12-guge copper wire in typicl residentil building hs cross-sectionl re of m 2. If it crries current of 10.0 A, wht is the drift speed of the electrons? Assume tht ech copper tom contributes one free electron to the current. The density of copper is 8.95 g/cm 3. Solution From the periodic tble of the elements in Appendix C, we find tht the molr mss of copper is 63.5 g/mol. Recll tht 1 mol of ny substnce contins Avogdro s number of toms ( ). Knowing the density of copper, we cn clculte the volume occupied by 63.5 g (1 mol) of copper: V m Becuse ech copper tom contributes one free electron to the current, we hve n electrons 7.09 cm electrons/m g 8.95 g/cm cm3 ( cm 3 /m 3 ) From Eqution 27.4, we find tht the drift speed is where q is the bsolute vlue of the chrge on ech electron. Thus, v d Exercise If copper wire crries current of 80.0 ma, how mny electrons flow pst given cross-section of the wire in 10.0 min? Answer I nqa 10.0 C/s ( m 3 )( C)( m 2 ) m/s v d electrons. I nqa

141 844 CHAPTER 27 Current nd Resistnce Exmple 27.1 shows tht typicl drift speeds re very low. For instnce, electrons trveling with speed of m/s would tke bout 68 min to trvel 1 m! In view of this, you might wonder why light turns on lmost instntneously when switch is thrown. In conductor, the electric field tht drives the free electrons trvels through the conductor with speed close to tht of light. Thus, when you flip on light switch, the messge for the electrons to strt moving through the wire (the electric field) reches them t speed on the order of 10 8 m/s RESISTANCE AND OHM S LAW In Chpter 24 we found tht no electric field cn exist inside conductor. However, this sttement is true only if the conductor is in sttic equilibrium. The purpose of this section is to describe wht hppens when the chrges in the conductor re llowed to move. Chrges moving in conductor produce current under the ction of n electric field, which is mintined by the connection of bttery cross the conductor. An electric field cn exist in the conductor becuse the chrges in this sitution re in motion tht is, this is nonelectrosttic sitution. Consider conductor of cross-sectionl re A crrying current I. The current density J in the conductor is defined s the current per unit re. Becuse the current I nqv d A, the current density is J I A nqv d (27.5) Current density Ohm s lw where J hs SI units of A/m 2. This expression is vlid only if the current density is uniform nd only if the surfce of cross-sectionl re A is perpendiculr to the direction of the current. In generl, the current density is vector quntity: J nqv d (27.6) From this eqution, we see tht current density, like current, is in the direction of chrge motion for positive chrge crriers nd opposite the direction of motion for negtive chrge crriers. A current density J nd n electric field E re estblished in conductor whenever potentil difference is mintined cross the conductor. If the potentil difference is constnt, then the current lso is constnt. In some mterils, the current density is proportionl to the electric field: J E (27.7) where the constnt of proportionlity is clled the conductivity of the conductor. 1 Mterils tht obey Eqution 27.7 re sid to follow Ohm s lw, nmed fter Georg Simon Ohm ( ). More specificlly, Ohm s lw sttes tht for mny mterils (including most metls), the rtio of the current density to the electric field is constnt tht is independent of the electric field producing the current. Mterils tht obey Ohm s lw nd hence demonstrte this simple reltionship between E nd J re sid to be ohmic. Experimentlly, it is found tht not ll mterils hve this property, however, nd mterils tht do not obey Ohm s lw re sid to 1 Do not confuse conductivity with surfce chrge density, for which the sme symbol is used.

142 27.2 Resistnce nd Ohm s Lw 845 be nonohmic. Ohm s lw is not fundmentl lw of nture but rther n empiricl reltionship vlid only for certin mterils. Quick Quiz 27.2 Suppose tht current-crrying ohmic metl wire hs cross-sectionl re tht grdully becomes smller from one end of the wire to the other. How do drift velocity, current density, nd electric field vry long the wire? Note tht the current must hve the sme vlue everywhere in the wire so tht chrge does not ccumulte t ny one point. We cn obtin form of Ohm s lw useful in prcticl pplictions by considering segment of stright wire of uniform cross-sectionl re A nd length, s shown in Figure A potentil difference V V b V is mintined cross the wire, creting in the wire n electric field nd current. If the field is ssumed to be uniform, the potentil difference is relted to the field through the reltionship 2 V E Therefore, we cn express the mgnitude of the current density in the wire s J E V Becuse J I/A, we cn write the potentil difference s V The quntity /A is clled the resistnce R of the conductor. We cn define the resistnce s the rtio of the potentil difference cross conductor to the current through the conductor: R A J A I V I (27.8) Resistnce of conductor From this result we see tht resistnce hs SI units of volts per mpere. One volt per mpere is defined to be 1 ohm (): 1 1 V 1 A (27.9) A I V b V E Figure 27.5 A uniform conductor of length nd cross-sectionl re A. A potentil difference V V b V mintined cross the conductor sets up n electric field E, nd this field produces current I tht is proportionl to the potentil difference. 2 This result follows from the definition of potentil difference: V b V b E ds E dx E 0

143 846 CHAPTER 27 Current nd Resistnce An ssortment of resistors used in electric circuits. Resistivity This expression shows tht if potentil difference of 1 V cross conductor cuses current of 1 A, the resistnce of the conductor is 1. For exmple, if n electricl pplince connected to 120-V source of potentil difference crries current of 6 A, its resistnce is 20. Eqution 27.8 solved for potentil difference (V I/A) explins prt of the chpter-opening puzzler: How cn bird perch on high-voltge power line without being electrocuted? Even though the potentil difference between the ground nd the wire might be hundreds of thousnds of volts, tht between the bird s feet (which is wht determines how much current flows through the bird) is very smll. The inverse of conductivity is resistivity 3 : 1 (27.10) where hs the units ohm-meters (m). We cn use this definition nd Eqution 27.8 to express the resistnce of uniform block of mteril s Resistnce of uniform conductor R A (27.11) Every ohmic mteril hs chrcteristic resistivity tht depends on the properties of the mteril nd on temperture. Additionlly, s you cn see from Eqution 27.11, the resistnce of smple depends on geometry s well s on resistivity. Tble 27.1 gives the resistivities of vriety of mterils t 20 C. Note the enormous rnge, from very low vlues for good conductors such s copper nd silver, to very high vlues for good insultors such s glss nd rubber. An idel conductor would hve zero resistivity, nd n idel insultor would hve infinite resistivity. Eqution shows tht the resistnce of given cylindricl conductor is proportionl to its length nd inversely proportionl to its cross-sectionl re. If the length of wire is doubled, then its resistnce doubles. If its cross-sectionl re is doubled, then its resistnce decreses by one hlf. The sitution is nlogous to the flow of liquid through pipe. As the pipe s length is incresed, the 3 Do not confuse resistivity with mss density or chrge density, for which the sme symbol is used.

144 27.2 Resistnce nd Ohm s Lw 847 TABLE 27.1 Resistivities nd Temperture Coefficients of Resistivity for Vrious Mterils Resistivity Temperture Mteril ( m) Coefficient [(C) 1 ] Silver Copper Gold Aluminum Tungsten Iron Pltinum Led Nichrome b Crbon Germnium Silicon Glss to Hrd rubber Sulfur Qurtz (fused) All vlues t 20 C. b A nickelchromium lloy commonly used in heting elements. resistnce to flow increses. As the pipe s cross-sectionl re is incresed, more liquid crosses given cross-section of the pipe per unit time. Thus, more liquid flows for the sme pressure differentil pplied to the pipe, nd the resistnce to flow decreses. Most electric circuits use devices clled resistors to control the current level in the vrious prts of the circuit. Two common types of resistors re the composition resistor, which contins crbon, nd the wire-wound resistor, which consists of coil of wire. Resistors vlues in ohms re normlly indicted by color-coding, s shown in Figure 27.6 nd Tble Ohmic mterils hve liner current potentil difference reltionship over brod rnge of pplied potentil differences (Fig. 27.7). The slope of the I-versus-V curve in the liner region yields vlue for 1/R. Nonohmic mterils Figure 27.6 The colored bnds on resistor represent code for determining resistnce. The first two colors give the first two digits in the resistnce vlue. The third color represents the power of ten for the multiplier of the resistnce vlue. The lst color is the tolernce of the resistnce vlue. As n exmple, the four colors on the circled resistors re red ( 2), blck ( 0), ornge ( 10 3 ), nd gold ( 5%), nd so the resistnce vlue is k with tolernce vlue of 5% 1k. (The vlues for the colors re from Tble 27.2.)

145 848 CHAPTER 27 Current nd Resistnce TABLE 27.2 Color Coding for Resistors Color Number Multiplier Tolernce Blck 0 1 Brown Red Ornge Yellow Green Blue Violet Gry White Gold % Silver % Colorless 20% I I Slope = 1 R V V Figure 27.7 () () The currentpotentil difference curve for n ohmic mteril. The curve is liner, nd the slope is equl to the inverse of the resistnce of the conductor. (b) A nonliner currentpotentil difference curve for semiconducting diode. This device does not obey Ohm s lw. (b) hve nonliner current potentil difference reltionship. One common semiconducting device tht hs nonliner I-versus-V chrcteristics is the junction diode (Fig. 27.7b). The resistnce of this device is low for currents in one direction (positive V ) nd high for currents in the reverse direction (negtive V ). In fct, most modern electronic devices, such s trnsistors, hve nonliner current potentil difference reltionships; their proper opertion depends on the prticulr wy in which they violte Ohm s lw. Quick Quiz 27.3 Wht does the slope of the curved line in Figure 27.7b represent? Quick Quiz 27.4 Your boss sks you to design n utomobile bttery jumper cble tht hs low resistnce. In view of Eqution 27.11, wht fctors would you consider in your design?

146 27.2 Resistnce nd Ohm s Lw 849 EXAMPLE 27.2 The Resistnce of Conductor Clculte the resistnce of n luminum cylinder tht is 10.0 cm long nd hs cross-sectionl re of m 2. Repet the clcultion for cylinder of the sme dimensions nd mde of glss hving resistivity of m. Solution From Eqution nd Tble 27.1, we cn clculte the resistnce of the luminum cylinder s follows: R A ( m) m m 2 ties, the resistnce of identiclly shped cylinders of luminum nd glss differ widely. The resistnce of the glss cylinder is 18 orders of mgnitude greter thn tht of the luminum cylinder Similrly, for glss we find tht R A ( m) m m As you might guess from the lrge difference in resistivi- Electricl insultors on telephone poles re often mde of glss becuse of its low electricl conductivity. EXAMPLE 27.3 The Resistnce of Nichrome Wire () Clculte the resistnce per unit length of 22-guge Nichrome wire, which hs rdius of mm. Solution The cross-sectionl re of this wire is A r 2 ( m) m 2 The resistivity of Nichrome is m (see Tble 27.1). Thus, we cn use Eqution to find the resistnce per unit length: (b) If potentil difference of 10 V is mintined cross 1.0-m length of the Nichrome wire, wht is the current in the wire? Solution R A m m 2 Becuse 1.0-m length of this wire hs resistnce of 4.6, Eqution 27.8 gives I V R 10 V A 4.6 /m Note from Tble 27.1 tht the resistivity of Nichrome wire is bout 100 times tht of copper. A copper wire of the sme rdius would hve resistnce per unit length of only /m. A 1.0-m length of copper wire of the sme rdius would crry the sme current (2.2 A) with n pplied potentil difference of only 0.11 V. Becuse of its high resistivity nd its resistnce to oxidtion, Nichrome is often used for heting elements in tosters, irons, nd electric heters. Exercise Wht is the resistnce of 6.0-m length of 22- guge Nichrome wire? How much current does the wire crry when connected to 120-V source of potentil difference? Answer 28 ; 4.3 A. Exercise Clculte the current density nd electric field in the wire when it crries current of 2.2 A. Answer A/m 2 ; 10 N/C. EXAMPLE 27.4 The Rdil Resistnce of Coxil Cble Coxil cbles re used extensively for cble television nd other electronic pplictions. A coxil cble consists of two cylindricl conductors. The gp between the conductors is completely filled with silicon, s shown in Figure 27.8, nd current lekge through the silicon is unwnted. (The cble is designed to conduct current long its length.) The rdius

147 850 CHAPTER 27 Current nd Resistnce of the inner conductor is cm, the rdius of the outer one is b 1.75 cm, nd the length of the cble is L 15.0 cm. Clculte the resistnce of the silicon between the two conductors. Solution In this type of problem, we must divide the object whose resistnce we re clculting into concentric elements of infinitesiml thickness dr (Fig. 27.8b). We strt by using the differentil form of Eqution 27.11, replcing with r for the distnce vrible: dr where dr is the resistnce of n element of silicon of thickness dr nd surfce re A. In this exmple, we tke s our representtive concentric element hollow silicon cylinder of rdius r, thickness dr, nd length L, s shown in Figure Any current tht psses from the inner conductor to the outer one must pss rdilly through this concentric element, nd the re through which this current psses is A 2rL. (This is the curved surfce re circumference multiplied by length of our hollow silicon cylinder of thickness dr.) Hence, we cn write the resistnce of our hollow cylinder of silicon s dr/a, Becuse we wish to know the totl resistnce cross the entire thickness of the silicon, we must integrte this expression from r to r b : Substituting in the vlues given, nd using 640 m for silicon, we obtin Exercise If potentil difference of 12.0 V is pplied between the inner nd outer conductors, wht is the vlue of the totl current tht psses between them? Answer R R b 640 m 2(0.150 m) ln 14.1 ma. dr dr 2L b 2rL dr dr r 1.75 cm cm 2L ln b 851 L Silicon dr Current direction r b Inner conductor Figure 27.8 Outer conductor () End view (b) A coxil cble. () Silicon fills the gp between the two conductors. (b) End view, showing current lekge A MODEL FOR ELECTRICAL CONDUCTION In this section we describe clssicl model of electricl conduction in metls tht ws first proposed by Pul Drude in This model leds to Ohm s lw nd shows tht resistivity cn be relted to the motion of electrons in metls. Although the Drude model described here does hve limittions, it nevertheless introduces concepts tht re still pplied in more elborte tretments. Consider conductor s regulr rry of toms plus collection of free electrons, which re sometimes clled conduction electrons. The conduction electrons, lthough bound to their respective toms when the toms re not prt of solid, gin mobility when the free toms condense into solid. In the bsence of n electric field, the conduction electrons move in rndom directions through the con-

148 27.3 A Model for Electricl Conduction 851 ductor with verge speeds of the order of 10 6 m/s. The sitution is similr to the motion of gs molecules confined in vessel. In fct, some scientists refer to conduction electrons in metl s n electron gs. There is no current through the conductor in the bsence of n electric field becuse the drift velocity of the free electrons is zero. Tht is, on the verge, just s mny electrons move in one direction s in the opposite direction, nd so there is no net flow of chrge. This sitution chnges when n electric field is pplied. Now, in ddition to undergoing the rndom motion just described, the free electrons drift slowly in direction opposite tht of the electric field, with n verge drift speed v d tht is much smller (typiclly 10 4 m/s) thn their verge speed between collisions (typiclly 10 6 m/s). Figure 27.9 provides crude description of the motion of free electrons in conductor. In the bsence of n electric field, there is no net displcement fter mny collisions (Fig. 27.9). An electric field E modifies the rndom motion nd cuses the electrons to drift in direction opposite tht of E (Fig. 27.9b). The slight curvture in the pths shown in Figure 27.9b results from the ccelertion of the electrons between collisions, which is cused by the pplied field. In our model, we ssume tht the motion of n electron fter collision is independent of its motion before the collision. We lso ssume tht the excess energy cquired by the electrons in the electric field is lost to the toms of the conductor when the electrons nd toms collide. The energy given up to the toms increses their vibrtionl energy, nd this cuses the temperture of the conductor to increse. The temperture increse of conductor due to resistnce is utilized in electric tosters nd other fmilir pplinces. We re now in position to derive n expression for the drift velocity. When free electron of mss m e nd chrge q (e) is subjected to n electric field E, it experiences force F qe. Becuse F m e, we conclude tht the ccelertion of the electron is qe (27.12) m e This ccelertion, which occurs for only short time between collisions, enbles the electron to cquire smll drift velocity. If t is the time since the lst collision nd v i is the electron s initil velocity the instnt fter tht collision, then the velocity of the electron fter time t is v f v i t v i qe t (27.13) m e We now tke the verge vlue of v f over ll possible times t nd ll possible vlues of v i. If we ssume tht the initil velocities re rndomly distributed over ll possible vlues, we see tht the verge vlue of v i is zero. The term (qe/m e )t is the velocity dded by the field during one trip between toms. If the electron strts with zero velocity, then the verge vlue of the second term of Eqution is (qe/m e ), where is the verge time intervl between successive collisions. Becuse the verge vlue of v f is equl to the drift velocity, 4 we hve v f v d qe m e (27.14) Figure 27.9 () E (b) () A schemtic digrm of the rndom motion of two chrge crriers in conductor in the bsence of n electric field. The drift velocity is zero. (b) The motion of the chrge crriers in conductor in the presence of n electric field. Note tht the rndom motion is modified by the field, nd the chrge crriers hve drift velocity. Drift velocity 4 Becuse the collision process is rndom, ech collision event is independent of wht hppened erlier. This is nlogous to the rndom process of throwing die. The probbility of rolling prticulr number on one throw is independent of the result of the previous throw. On verge, the prticulr number comes up every sixth throw, strting t ny rbitrry time.

149 852 CHAPTER 27 Current nd Resistnce Current density Conductivity Resistivity We cn relte this expression for drift velocity to the current in the conductor. Substituting Eqution into Eqution 27.6, we find tht the mgnitude of the current density is J nqv d nq 2 E (27.15) m e where n is the number of chrge crriers per unit volume. Compring this expression with Ohm s lw, J E, we obtin the following reltionships for conductivity nd resistivity: 1 nq 2 m e (27.16) (27.17) According to this clssicl model, conductivity nd resistivity do not depend on the strength of the electric field. This feture is chrcteristic of conductor obeying Ohm s lw. The verge time between collisions is relted to the verge distnce between collisions (tht is, the men free pth; see Section 21.7) nd the verge speed v through the expression v m e nq 2 (27.18) EXAMPLE 27.5 Electron Collisions in Wire () Using the dt nd results from Exmple 27.1 nd the clssicl model of electron conduction, estimte the verge time between collisions for electrons in household copper wiring. Solution From Eqution 27.17, we see tht where for copper nd the crrier density is n electrons/m 3 for the wire described in Exmple Substitution of these vlues into the expression bove gives m m e nq 2 ( kg) ( m 3 )( C) 2 ( m) (b) Assuming tht the verge speed for free electrons in copper is m/s nd using the result from prt (), clculte the men free pth for electrons in copper. Solution s v ( m/s)( s) m which is equivlent to 40 nm (compred with tomic spcings of bout 0.2 nm). Thus, lthough the time between collisions is very short, n electron in the wire trvels bout 200 tomic spcings between collisions. Although this clssicl model of conduction is consistent with Ohm s lw, it is not stisfctory for explining some importnt phenomen. For exmple, clssicl vlues for v clculted on the bsis of n idel-gs model (see Section 21.6) re smller thn the true vlues by bout fctor of ten. Furthermore, if we substitute / v for in Eqution nd rerrnge terms so tht v ppers in the numertor, we find tht the resistivity is proportionl to v. According to the idel-gs model, v is proportionl to!t ; hence, it should lso be true tht. This is in disgreement with the fct tht, for pure metls, resistivity depends linerly on temperture. We re ble to ccount for the liner dependence only by using quntum mechnicl model, which we now describe briefly.!t

150 27.4 Resistnce nd Temperture 853 According to quntum mechnics, electrons hve wve-like properties. If the rry of toms in conductor is regulrly spced (tht is, it is periodic), then the wve-like chrcter of the electrons enbles them to move freely through the conductor, nd collision with n tom is unlikely. For n idelized conductor, no collisions would occur, the men free pth would be infinite, nd the resistivity would be zero. Electron wves re scttered only if the tomic rrngement is irregulr (not periodic) s result of, for exmple, structurl defects or impurities. At low tempertures, the resistivity of metls is dominted by scttering cused by collisions between electrons nd defects or impurities. At high tempertures, the resistivity is dominted by scttering cused by collisions between electrons nd toms of the conductor, which re continuously displced from the regulrly spced rry s result of therml gittion. The therml motion of the toms cuses the structure to be irregulr (compred with n tomic rry t rest), thereby reducing the electron s men free pth RESISTANCE AND TEMPERATURE Over limited temperture rnge, the resistivity of metl vries pproximtely linerly with temperture ccording to the expression 0[1 (T T 0 )] (27.19) where is the resistivity t some temperture T (in degrees Celsius), 0 is the resistivity t some reference temperture T 0 (usully tken to be 20 C), nd is the temperture coefficient of resistivity. From Eqution 27.19, we see tht the temperture coefficient of resistivity cn be expressed s (27.20) 0 T where 0 is the chnge in resistivity in the temperture intervl T T T 0. The temperture coefficients of resistivity for vrious mterils re given in Tble Note tht the unit for is degrees Celsius 1 [( C) 1 ]. Becuse resistnce is proportionl to resistivity (Eq ), we cn write the vrition of resistnce s R R 0 [1 (T T 0 )] (27.21) Use of this property enbles us to mke precise temperture mesurements, s shown in the following exmple. 1 Vrition of with temperture Temperture coefficient of resistivity EXAMPLE 27.6 A Pltinum Resistnce Thermometer A resistnce thermometer, which mesures temperture by mesuring the chnge in resistnce of conductor, is mde from pltinum nd hs resistnce of 50.0 t 20.0 C. When immersed in vessel contining melting indium, its resistnce increses to Clculte the melting point of the indium. Solution Solving Eqution for T nd using the vlue for pltinum given in Tble 27.1, we obtin T R R 0 R 0 Becuse T C, we find tht T, the temperture of the melting indium smple, is 157C [ (C) 1 ](50.0 ) 137C

151 854 CHAPTER 27 Current nd Resistnce ρ ρ 0 T T Figure Resistivity versus temperture for pure semiconductor, such s silicon or germnium. ρ 0 0 ρ Figure Resistivity versus temperture for metl such s copper. The curve is liner over wide rnge of tempertures, nd increses with incresing temperture. As T pproches bsolute zero (inset), the resistivity pproches finite vlue 0. T For metls like copper, resistivity is nerly proportionl to temperture, s shown in Figure However, nonliner region lwys exists t very low tempertures, nd the resistivity usully pproches some finite vlue s the temperture ners bsolute zero. This residul resistivity ner bsolute zero is cused primrily by the collision of electrons with impurities nd imperfections in the metl. In contrst, high-temperture resistivity (the liner region) is predominntly chrcterized by collisions between electrons nd metl toms. Notice tht three of the vlues in Tble 27.1 re negtive; this indictes tht the resistivity of these mterils decreses with incresing temperture (Fig ). This behvior is due to n increse in the density of chrge crriers t higher tempertures. Becuse the chrge crriers in semiconductor re often ssocited with impurity toms, the resistivity of these mterils is very sensitive to the type nd concentrtion of such impurities. We shll return to the study of semiconductors in Chpter 43 of the extended version of this text. Quick Quiz 27.5 When does lightbulb crry more current just fter it is turned on nd the glow of the metl filment is incresing, or fter it hs been on for few milliseconds nd the glow is stedy? Optionl Section 27.5 SUPERCONDUCTORS There is clss of metls nd compounds whose resistnce decreses to zero when they re below certin temperture T c, known s the criticl temperture. These mterils re known s superconductors. The resistnce temperture grph for superconductor follows tht of norml metl t tempertures bove T c (Fig ). When the temperture is t or below T c, the resistivity drops suddenly to zero. This phenomenon ws discovered in 1911 by the Dutch physicist Heike Kmerlingh-Onnes ( ) s he worked with mercury, which is superconductor below 4.2 K. Recent mesurements hve shown tht the resistivities of superconductors below their T c vlues re less thn m round times smller thn the resistivity of copper nd in prctice considered to be zero. Tody thousnds of superconductors re known, nd s Figure illustrtes, the criticl tempertures of recently discovered superconductors re substntilly higher thn initilly thought possible. Two kinds of superconductors re recognized. The more recently identified ones, such s YB 2 Cu 3 O 7, re essentilly cermics with high criticl tempertures, wheres superconducting mterils such

152 27.5 Superconductors 855 R(Ω) Hg T c T(K) Figure Resistnce versus temperture for smple of mercury (Hg). The grph follows tht of norml metl bove the criticl temperture T c. The resistnce drops to zero t T c, which is 4.2 K for mercury. s those observed by Kmerlingh-Onnes re metls. If room-temperture superconductor is ever identified, its impct on technology could be tremendous. The vlue of T c is sensitive to chemicl composition, pressure, nd moleculr structure. It is interesting to note tht copper, silver, nd gold, which re excellent conductors, do not exhibit superconductivity. A smll permnent mgnet levitted bove disk of the superconductor Y B 2 Cu 3 O 7, which is t 77 K. T c (K) Hg-B 2 C 2 Cu 2 O 8 δ Tl-B-C-Cu-O Bi-B-C-Cu-O Liquid O 2 90 YB 2 Cu 3 O 7 δ Liquid N L-Sr-Cu-O 40 Nb 3 Ge Liquid H Hg NbN L-B-Cu-O 10 Liquid He Yer of discovery Figure phenomenon. Evolution of the superconducting criticl temperture since the discovery of the

153 856 CHAPTER 27 Current nd Resistnce One of the truly remrkble fetures of superconductors is tht once current is set up in them, it persists without ny pplied potentil difference (becuse R 0). Stedy currents hve been observed to persist in superconducting loops for severl yers with no pprent decy! An importnt nd useful ppliction of superconductivity is in the development of superconducting mgnets, in which the mgnitudes of the mgnetic field re bout ten times greter thn those produced by the best norml electromgnets. Such superconducting mgnets re being considered s mens of storing energy. Superconducting mgnets re currently used in medicl mgnetic resonnce imging (MRI) units, which produce high-qulity imges of internl orgns without the need for excessive exposure of ptients to x-rys or other hrmful rdition. For further informtion on superconductivity, see Section b Power V Figure I A circuit consisting of resistor of resistnce R nd bttery hving potentil difference V cross its terminls. Positive chrge flows in the clockwise direction. Points nd d re grounded. R c d ELECTRICAL ENERGY AND POWER If bttery is used to estblish n electric current in conductor, the chemicl energy stored in the bttery is continuously trnsformed into kinetic energy of the chrge crriers. In the conductor, this kinetic energy is quickly lost s result of collisions between the chrge crriers nd the toms mking up the conductor, nd this leds to n increse in the temperture of the conductor. In other words, the chemicl energy stored in the bttery is continuously trnsformed to internl energy ssocited with the temperture of the conductor. Consider simple circuit consisting of bttery whose terminls re connected to resistor, s shown in Figure (Resistors re designted by the symbol.) Now imgine following positive quntity of chrge Q tht is moving clockwise round the circuit from point through the bttery nd resistor bck to point. Points nd d re grounded (ground is designted by the symbol ); tht is, we tke the electric potentil t these two points to be zero. As the chrge moves from to b through the bttery, its electric potentil energy U increses by n mount V Q (where V is the potentil difference between b nd ), while the chemicl potentil energy in the bttery decreses by the sme mount. (Recll from Eq tht U q V.) However, s the chrge moves from c to d through the resistor, it loses this electric potentil energy s it collides with toms in the resistor, thereby producing internl energy. If we neglect the resistnce of the connecting wires, no loss in energy occurs for pths bc nd d. When the chrge rrives t point, it must hve the sme electric potentil energy (zero) tht it hd t the strt. 5 Note tht becuse chrge cnnot build up t ny point, the current is the sme everywhere in the circuit. The rte t which the chrge Q loses potentil energy in going through the resistor is U t Q t V I V where I is the current in the circuit. In contrst, the chrge regins this energy when it psses through the bttery. Becuse the rte t which the chrge loses energy equls the power delivered to the resistor (which ppers s internl energy), we hve I V (27.22) 5 Note tht once the current reches its stedy-stte vlue, there is no chnge in the kinetic energy of the chrge crriers creting the current.

154 27.6 Electricl Energy nd Power 857 In this cse, the power is supplied to resistor by bttery. However, we cn use Eqution to determine the power trnsferred to ny device crrying current I nd hving potentil difference V between its terminls. Using Eqution nd the fct tht V IR for resistor, we cn express the power delivered to the resistor in the lterntive forms I 2 (V )2 R (27.23) R When I is expressed in mperes, V in volts, nd R in ohms, the SI unit of power is the wtt, s it ws in Chpter 7 in our discussion of mechnicl power. The power lost s internl energy in conductor of resistnce R is clled joule heting 6 ; this trnsformtion is lso often referred to s n I 2 R loss. A bttery, device tht supplies electricl energy, is clled either source of electromotive force or, more commonly, n emf source. The concept of emf is discussed in greter detil in Chpter 28. (The phrse electromotive force is n unfortunte choice becuse it describes not force but rther potentil difference in volts.) When the internl resistnce of the bttery is neglected, the potentil difference between points nd b in Figure is equl to the emf of the bttery tht is, V V b V. This being true, we cn stte tht the current in the circuit is I V/R /R. Becuse V, the power supplied by the emf source cn be expressed s I, which equls the power delivered to the resistor, I 2 R. When trnsporting electricl energy through power lines, such s those shown in Figure 27.15, utility compnies seek to minimize the power trnsformed to internl energy in the lines nd mximize the energy delivered to the consumer. Becuse I V, the sme mount of power cn be trnsported either t high currents nd low potentil differences or t low currents nd high potentil differences. Utility compnies choose to trnsport electricl energy t low currents nd high potentil differences primrily for economic resons. Copper wire is very expensive, nd so it is cheper to use high-resistnce wire (tht is, wire hving smll cross-sectionl re; see Eq ). Thus, in the expression for the power delivered to resistor, I 2 R, the resistnce of the wire is fixed t reltively high vlue for economic considertions. The I 2 R loss cn be reduced by keeping the current I s low s possible. In some instnces, power is trnsported t potentil differences s gret s 765 kv. Once the electricity reches your city, the potentil difference is usully reduced to 4 kv by device clled trnsformer. Another trnsformer drops the potentil difference to 240 V before the electricity finlly reches your home. Of course, ech time the potentil difference decreses, the current increses by the sme fctor, nd the power remins the sme. We shll discuss trnsformers in greter detil in Chpter 33. Power delivered to resistor Figure Power compnies trnsfer electricl energy t high potentil differences. Quick Quiz 27.6 The sme potentil difference is pplied to the two lightbulbs shown in Figure Which one of the following sttements is true? () The 30-W bulb crries the greter current nd hs the higher resistnce. (b) The 30-W bulb crries the greter current, but the 60-W bulb hs the higher resistnce. QuickLb If you hve ccess to n ohmmeter, verify your nswer to Quick Quiz 27.6 by testing the resistnce of few lightbulbs. 6 It is clled joule heting even though the process of het does not occur. This is nother exmple of incorrect usge of the word het tht hs become entrenched in our lnguge.

155 858 CHAPTER 27 Current nd Resistnce Figure These lightbulbs operte t their rted power only when they re connected to 120-V source. (c) The 30-W bulb hs the higher resistnce, but the 60-W bulb crries the greter current. (d) The 60-W bulb crries the greter current nd hs the higher resistnce. QuickLb From the lbels on household pplinces such s hir dryers, televisions, nd stereos, estimte the nnul cost of operting them. Quick Quiz 27.7 For the two lightbulbs shown in Figure 27.17, rnk the current vlues t points through f, from gretest to lest. 30 W e f 60 W c d V b Figure Two lightbulbs connected cross the sme potentil difference. The bulbs operte t their rted power only if they re connected to 120-V bttery. EXAMPLE 27.7 Power in n Electric Heter An electric heter is constructed by pplying potentil difference of 120 V to Nichrome wire tht hs totl resistnce of Find the current crried by the wire nd the power rting of the heter. Solution Becuse V IR, we hve I V R 120 V A We cn find the power rting using the expression I 2 R : I 2 R (15.0 A) 2 (8.00 ) 1.80 kw If we doubled the pplied potentil difference, the current would double but the power would qudruple becuse (V ) 2 /R.

156 27.6 Electricl Energy nd Power 859 EXAMPLE 27.8 The Cost of Mking Dinner Estimte the cost of cooking turkey for 4 h in n oven tht opertes continuously t 20.0 A nd 240 V. Solution The power used by the oven is I V (20.0 A)(240 V) W 4.80 kw Becuse the energy consumed equls power time, the mount of energy for which you must py is Energy t (4.80 kw)(4 h) 19.2 kwh If the energy is purchsed t n estimted price of 8.00 per kilowtt hour, the cost is Cost (19.2 kwh)($0.080/kwh) $1.54 Demnds on our dwindling energy supplies hve mde it necessry for us to be wre of the energy requirements of our electricl devices. Every electricl pplince crries lbel tht contins the informtion you need to clculte the pplince s power requirements. In mny cses, the power consumption in wtts is stted directly, s it is on lightbulb. In other cses, the mount of current used by the device nd the potentil difference t which it opertes re given. This informtion nd Eqution re sufficient for clculting the operting cost of ny electricl device. Exercise Wht does it cost to operte 100-W lightbulb for 24 h if the power compny chrges $0.08/kWh? Answer $0.19. EXAMPLE 27.9 Current in n Electron Bem In certin prticle ccelertor, electrons emerge with n energy of 40.0 MeV (1 MeV J). The electrons emerge not in stedy strem but rther in pulses t the rte of 250 pulses/s. This corresponds to time between pulses of 4.00 ms (Fig ). Ech pulse hs durtion of 200 ns, nd the electrons in the pulse constitute current of 250 ma. The current is zero between pulses. () How mny electrons re delivered by the ccelertor per pulse? Solution We use Eqution 27.2 in the form dq I dt nd integrte to find the chrge per pulse. While the pulse is on, the current is constnt; thus, Q pulse I dt It ( A)( s) C Dividing this quntity of chrge per pulse by the electronic chrge gives the number of electrons per pulse: Electrons per pulse (b) Wht is the verge current per pulse delivered by the ccelertor? Solution Averge current is given by Eqution 27.1, I v Q /t. Becuse the time intervl between pulses is 4.00 ms, nd becuse we know the chrge per pulse from prt (), we obtin I v Q pulse t C/pulse C/electron electrons/pulse C s 12.5 A This represents only 0.005% of the pek current, which is 250 ma. I 4.00 ms s t (s) Figure electrons. Current versus time for pulsed bem of

157 860 CHAPTER 27 Current nd Resistnce (c) Wht is the mximum power delivered by the electron bem? Solution By definition, power is energy delivered per unit time. Thus, the mximum power is equl to the energy delivered by pulse divided by the pulse durtion: E t ( electrons/pulse)(40.0 MeV/electron) s/pulse ( MeV/s)( J/MeV ) W 10.0 MW We could lso compute this power directly. We ssume tht ech electron hd zero energy before being ccelerted. Thus, by definition, ech electron must hve gone through potentil difference of 40.0 MV to cquire finl energy of 40.0 MeV. Hence, we hve I V ( A)( V) 10.0 MW SUMMARY The electric current I in conductor is defined s I dq (27.2) dt where dq is the chrge tht psses through cross-section of the conductor in time dt. The SI unit of current is the mpere (A), where 1 A 1 C/s. The verge current in conductor is relted to the motion of the chrge crriers through the reltionship I v nqv d A (27.4) where n is the density of chrge crriers, q is the chrge on ech crrier, v d is the drift speed, nd A is the cross-sectionl re of the conductor. The mgnitude of the current density J in conductor is the current per unit re: J I A nqv d (27.5) The current density in conductor is proportionl to the electric field ccording to the expression J E (27.7) The proportionlity constnt is clled the conductivity of the mteril of which the conductor is mde. The inverse of is known s resistivity ( 1/). Eqution 27.7 is known s Ohm s lw, nd mteril is sid to obey this lw if the rtio of its current density J to its pplied electric field E is constnt tht is independent of the pplied field. The resistnce R of conductor is defined either in terms of the length of the conductor or in terms of the potentil difference cross it: R A V I (27.8) where is the length of the conductor, is the conductivity of the mteril of which it is mde, A is its cross-sectionl re, V is the potentil difference cross it, nd I is the current it crries.

158 Questions 861 The SI unit of resistnce is volts per mpere, which is defined to be 1 ohm (); tht is, 1 1 V/A. If the resistnce is independent of the pplied potentil difference, the conductor obeys Ohm s lw. In clssicl model of electricl conduction in metls, the electrons re treted s molecules of gs. In the bsence of n electric field, the verge velocity of the electrons is zero. When n electric field is pplied, the electrons move (on the verge) with drift velocity v d tht is opposite the electric field nd given by the expression v d qe (27.14) m e where is the verge time between electrontom collisions, m e is the mss of the electron, nd q is its chrge. According to this model, the resistivity of the metl is (27.17) nq 2 where n is the number of free electrons per unit volume. The resistivity of conductor vries pproximtely linerly with temperture ccording to the expression 0[1 (T T 0 )] (27.19) where is the temperture coefficient of resistivity nd 0 is the resistivity t some reference temperture T 0. If potentil difference V is mintined cross resistor, the power, or rte t which energy is supplied to the resistor, is I V (27.22) Becuse the potentil difference cross resistor is given by V IR, we cn express the power delivered to resistor in the form I 2 (V )2 R (27.23) R The electricl energy supplied to resistor ppers in the form of internl energy in the resistor. m e QUESTIONS 1. Newspper rticles often contin sttements such s volts of electricity surged through the victim s body. Wht is wrong with this sttement? 2. Wht is the difference between resistnce nd resistivity? 3. Two wires A nd B of circulr cross-section re mde of the sme metl nd hve equl lengths, but the resistnce of wire A is three times greter thn tht of wire B. Wht is the rtio of their cross-sectionl res? How do their rdii compre? 4. Wht is required in order to mintin stedy current in conductor? 5. Do ll conductors obey Ohm s lw? Give exmples to justify your nswer. 6. When the voltge cross certin conductor is doubled, the current is observed to increse by fctor of three. Wht cn you conclude bout the conductor? 7. In the wter nlogy of n electric circuit, wht corresponds to the power supply, resistor, chrge, nd potentil difference? 8. Why might good electricl conductor lso be good therml conductor? 9. On the bsis of the tomic theory of mtter, explin why the resistnce of mteril should increse s its temperture increses. 10. How does the resistnce for copper nd silicon chnge with temperture? Why re the behviors of these two mterils different? 11. Explin how current cn persist in superconductor in the bsence of ny pplied voltge. 12. Wht single experimentl requirement mkes superconducting devices expensive to operte? In principle, cn this limittion be overcome?

159 862 CHAPTER 27 Current nd Resistnce 13. Wht would hppen to the drift velocity of the electrons in wire nd to the current in the wire if the electrons could move freely without resistnce through the wire? 14. If chrges flow very slowly through metl, why does it not require severl hours for light to turn on when you throw switch? 15. In conductor, the electric field tht drives the electrons through the conductor propgtes with speed tht is lmost the sme s the speed of light, even though the drift velocity of the electrons is very smll. Explin how these cn both be true. Does given electron move from one end of the conductor to the other? 16. Two conductors of the sme length nd rdius re connected cross the sme potentil difference. One conductor hs twice the resistnce of the other. To which conductor is more power delivered? 17. Cr btteries re often rted in mpere-hours. Does this designte the mount of current, power, energy, or chrge tht cn be drwn from the bttery? 18. If you were to design n electric heter using Nichrome wire s the heting element, wht prmeters of the wire could you vry to meet specific power output, such s W? 19. Consider the following typicl monthly utility rte structure: $2.00 for the first 16 kwh, 8.00 /kwh for the next 34 kwh, 6.50 /kwh for the next 50 kwh, 5.00 /kwh for the next 100 kwh, 4.00 /kwh for the next 200 kwh, nd 3.50 /kwh for ll kilowtt-hours in excess of 400 kwh. On the bsis of these rtes, determine the mount chrged for 327 kwh. PROBLEMS 1, 2, 3 = strightforwrd, intermedite, chllenging = full solution vilble in the Student Solutions Mnul nd Study Guide WEB = solution posted t = Computer useful in solving problem = Interctive Physics = pired numericl/symbolic problems Section 27.1 Electric Current 1. In prticulr cthode ry tube, the mesured bem current is 30.0 A. How mny electrons strike the tube screen every 40.0 s? 2. A tepot with surfce re of 700 cm 2 is to be silver plted. It is ttched to the negtive electrode of n electrolytic cell contining silver nitrte (Ag NO 3 ). If the cell is powered by 12.0-V bttery nd hs resistnce of 1.80, how long does it tke for mm lyer of silver to build up on the tepot? (The density of silver is kg/m 3.) 3. Suppose tht the current through conductor decreses exponentilly with time ccording to the expression I(t) I 0 e t/, where I 0 is the initil current (t t 0) nd is constnt hving dimensions of time. Consider fixed observtion point within the conductor. () How much chrge psses this point between t 0 nd t (b) How much chrge psses this point between t 0 nd t 10? (c) How much chrge psses this point between t 0 nd t? 4. In the Bohr model of the hydrogen tom, n electron in the lowest energy stte follows circulr pth t distnce of m from the proton. () Show tht the speed of the electron is m/s. (b) Wht is the effective current ssocited with this orbiting electron? 5. A smll sphere tht crries chrge of 8.00 nc is whirled in circle t the end of n insulting string. The ngulr frequency of rottion is 100 rd/s. Wht verge current does this rotting chrge represent? WEB? 6. A smll sphere tht crries chrge q is whirled in circle t the end of n insulting string. The ngulr frequency of rottion is. Wht verge current does this rotting chrge represent? 7. The quntity of chrge q (in coulombs) pssing through surfce of re 2.00 cm 2 vries with time ccording to the eqution q 4.00t t 6.00, where t is in seconds. () Wht is the instntneous current through the surfce t t 1.00 s? (b) Wht is the vlue of the current density? 8. An electric current is given by the expression I(t) 100 sin(120t), where I is in mperes nd t is in seconds. Wht is the totl chrge crried by the current from t 0 to t 1/240 s? 9. Figure P27.9 represents section of circulr conductor of nonuniform dimeter crrying current of 5.00 A. The rdius of cross-section A 1 is cm. () Wht is the mgnitude of the current density cross A 1? (b) If the current density cross A 2 is one-fourth the vlue cross A 1, wht is the rdius of the conductor t A 2? A 1 Figure P27.9 I A 2

160 Problems 863 WEB 10. A Vn de Grff genertor produces bem of 2.00-MeV deuterons, which re hevy hydrogen nuclei contining proton nd neutron. () If the bem current is 10.0 A, how fr prt re the deuterons? (b) Is their electrosttic repulsion fctor in bem stbility? Explin. 11. The electron bem emerging from certin highenergy electron ccelertor hs circulr cross-section of rdius 1.00 mm. () If the bem current is 8.00 A, wht is the current density in the bem, ssuming tht it is uniform throughout? (b) The speed of the electrons is so close to the speed of light tht their speed cn be tken s c m/s with negligible error. Find the electron density in the bem. (c) How long does it tke for Avogdro s number of electrons to emerge from the ccelertor? 12. An luminum wire hving cross-sectionl re of m 2 crries current of 5.00 A. Find the drift speed of the electrons in the wire. The density of luminum is 2.70 g/cm 3. (Assume tht one electron is supplied by ech tom.) Section 27.2 Resistnce nd Ohm s Lw 13. A lightbulb hs resistnce of 240 when operting t voltge of 120 V. Wht is the current through the lightbulb? 14. A resistor is constructed of crbon rod tht hs uniform cross-sectionl re of 5.00 mm 2. When potentil difference of 15.0 V is pplied cross the ends of the rod, there is current of A in the rod. Find () the resistnce of the rod nd (b) the rod s length. 15. A V potentil difference is mintined cross 1.50-m length of tungsten wire tht hs cross-sectionl re of mm 2. Wht is the current in the wire? 16. A conductor of uniform rdius 1.20 cm crries current of 3.00 A produced by n electric field of 120 V/m. Wht is the resistivity of the mteril? 17. Suppose tht you wish to fbricte uniform wire out of 1.00 g of copper. If the wire is to hve resistnce of R 0.500, nd if ll of the copper is to be used, wht will be () the length nd (b) the dimeter of this wire? 18. () Mke n order-of-mgnitude estimte of the resistnce between the ends of rubber bnd. (b) Mke n order-of-mgnitude estimte of the resistnce between the heds nd tils sides of penny. In ech cse, stte wht quntities you tke s dt nd the vlues you mesure or estimte for them. (c) Wht would be the order of mgnitude of the current tht ech crries if it were connected cross 120-V power supply? (WARNING! Do not try this t home!) 19. A solid cube of silver (density 10.5 g/cm 3 ) hs mss of 90.0 g. () Wht is the resistnce between opposite fces of the cube? (b) If there is one conduction electron for ech silver tom, wht is the verge drift speed of electrons when potentil difference of V is pplied to opposite fces? (The WEB tomic number of silver is 47, nd its molr mss is g/mol.) 20. A metl wire of resistnce R is cut into three equl pieces tht re then connected side by side to form new wire whose length is equl to one-third the originl length. Wht is the resistnce of this new wire? 21. A wire with resistnce R is lengthened to 1.25 times its originl length by being pulled through smll hole. Find the resistnce of the wire fter it hs been stretched. 22. Aluminum nd copper wires of equl length re found to hve the sme resistnce. Wht is the rtio of their rdii? 23. A current density of A/m 2 exists in the tmosphere where the electric field (due to chrged thunderclouds in the vicinity) is 100 V/m. Clculte the electricl conductivity of the Erth s tmosphere in this region. 24. The rod in Figure P27.24 (not drwn to scle) is mde of two mterils. Both hve squre cross section of 3.00 mm on side. The first mteril hs resistivity of m nd is 25.0 cm long, while the second mteril hs resistivity of m nd is 40.0 cm long. Wht is the resistnce between the ends of the rod? Section 27.3 Figure P27.24 A Model for Electricl Conduction 25. If the drift velocity of free electrons in copper wire is m/s, wht is the electric field in the conductor? 26. If the current crried by conductor is doubled, wht hppens to the () chrge crrier density? (b) current density? (c) electron drift velocity? (d) verge time between collisions? 27. Use dt from Exmple 27.1 to clculte the collision men free pth of electrons in copper, ssuming tht the verge therml speed of conduction electrons is m/s. Section cm 40.0 cm Resistnce nd Temperture 28. While tking photogrphs in Deth Vlley on dy when the temperture is 58.0 C, Bill Hiker finds tht certin voltge pplied to copper wire produces current of A. Bill then trvels to Antrctic nd pplies the sme voltge to the sme wire. Wht current does he register there if the temperture is 88.0 C? Assume tht no chnge occurs in the wire s shpe nd size. 29. A certin lightbulb hs tungsten filment with resistnce of 19.0 when cold nd of 140 when hot. Assuming tht Eqution cn be used over the lrge

161 864 CHAPTER 27 Current nd Resistnce WEB temperture rnge involved here, find the temperture of the filment when hot. (Assume n initil temperture of 20.0 C.) 30. A crbon wire nd Nichrome wire re connected in series. If the combintion hs resistnce of 10.0 k t 0 C, wht is the resistnce of ech wire t 0 C such tht the resistnce of the combintion does not chnge with temperture? (Note tht the equivlent resistnce of two resistors in series is the sum of their resistnces.) 31. An luminum wire with dimeter of mm hs uniform electric field with mgnitude of V/m imposed long its entire length. The temperture of the wire is 50.0 C. Assume one free electron per tom. () Using the informtion given in Tble 27.1, determine the resistivity. (b) Wht is the current density in the wire? (c) Wht is the totl current in the wire? (d) Wht is the drift speed of the conduction electrons? (e) Wht potentil difference must exist between the ends of 2.00-m length of the wire if the stted electric field is to be produced? 32. Review Problem. An luminum rod hs resistnce of t 20.0 C. Clculte the resistnce of the rod t 120 C by ccounting for the chnges in both the resistivity nd the dimensions of the rod. 33. Wht is the frctionl chnge in the resistnce of n iron filment when its temperture chnges from 25.0 C to 50.0 C? 34. The resistnce of pltinum wire is to be clibrted for low-temperture mesurements. A pltinum wire with resistnce of 1.00 t 20.0 C is immersed in liquid nitrogen t 77 K ( 196 C). If the temperture response of the pltinum wire is liner, wht is the expected resistnce of the pltinum wire t 196 C? (pltinum / C) 35. The temperture of tungsten smple is rised while copper smple is mintined t 20 C. At wht temperture will the resistivity of the tungsten smple be four times tht of the copper smple? 36. A segment of Nichrome wire is initilly t 20.0 C. Using the dt from Tble 27.1, clculte the temperture to which the wire must be heted if its resistnce is to be doubled. Section 27.6 Electricl Energy nd Power 37. A toster is rted t 600 W when connected to 120-V source. Wht current does the toster crry, nd wht is its resistnce? 38. In hydroelectric instlltion, turbine delivers hp to genertor, which in turn converts 80.0% of the mechnicl energy into electricl energy. Under these conditions, wht current does the genertor deliver t terminl potentil difference of V? 39. Review Problem. Wht is the required resistnce of n immersion heter tht increses the temperture of 1.50 kg of wter from 10.0 C to 50.0 C in 10.0 min while operting t 110 V? 40. Review Problem. Wht is the required resistnce of n immersion heter tht increses the temperture of mss m of liquid wter from T 1 to T 2 in time t while operting t voltge V? 41. Suppose tht voltge surge produces 140 V for moment. By wht percentge does the power output of 120-V, 100-W lightbulb increse? (Assume tht its resistnce does not chnge.) 42. A 500-W heting coil designed to operte from 110 V is mde of Nichrome wire mm in dimeter. () Assuming tht the resistivity of the Nichrome remins constnt t its 20.0 C vlue, find the length of wire used. (b) Now consider the vrition of resistivity with temperture. Wht power does the coil of prt () ctully deliver when it is heted to C? 43. A coil of Nichrome wire is 25.0 m long. The wire hs dimeter of mm nd is t 20.0 C. If it crries current of A, wht re () the mgnitude of the electric field in the wire nd (b) the power delivered to it? (c) If the temperture is incresed to 340 C nd the potentil difference cross the wire remins constnt, wht is the power delivered? 44. Btteries re rted in terms of mpere-hours (A h): For exmple, bttery tht cn produce current of 2.00 A for 3.00 h is rted t 6.00 A h. () Wht is the totl energy, in kilowtt-hours, stored in 12.0-V bttery rted t 55.0 A h? (b) At rte of $ per kilowtt-hour, wht is the vlue of the electricity produced by this bttery? 45. A 10.0-V bttery is connected to 120- resistor. Neglecting the internl resistnce of the bttery, clculte the power delivered to the resistor. 46. It is estimted tht ech person in the United Sttes (popultion 270 million) hs one electric clock, nd tht ech clock uses energy t rte of 2.50 W. To supply this energy, bout how mny metric tons of col re burned per hour in col-fired electricity generting plnts tht re, on verge, 25.0% efficient? (The het of combustion for col is 33.0 MJ/kg.) 47. Compute the cost per dy of operting lmp tht drws 1.70 A from 110-V line if the cost of electricl energy is $ /kWh. 48. Review Problem. The heting element of coffeemker opertes t 120 V nd crries current of 2.00 A. Assuming tht ll of the energy trnsferred from the heting element is bsorbed by the wter, clculte how long it tkes to het kg of wter from room temperture (23.0 C) to the boiling point. 49. A certin toster hs heting element mde of Nichrome resistnce wire. When the toster is first connected to 120-V source of potentil difference (nd the wire is t temperture of 20.0 C), the initil current is 1.80 A. However, the current begins to decrese s the resistive element wrms up. When the toster hs reched its finl operting temperture, the current hs dropped to 1.53 A. () Find the power the toster con-

162 Problems 865 WEB sumes when it is t its operting temperture. (b) Wht is the finl temperture of the heting element? 50. To het room hving ceilings 8.0 ft high, bout 10.0 W of electric power re required per squre foot. At cost of $ /kWh, how much does it cost per dy to use electricity to het room mesuring 10.0 ft 15.0 ft? 51. Estimte the cost of one person s routine use of hir dryer for 1 yr. If you do not use blow dryer yourself, observe or interview someone who does. Stte the quntities you estimte nd their vlues. ADDITIONAL PROBLEMS 52. One lightbulb is mrked 25 W 120 V, nd nother 100 W 120 V ; this mens tht ech bulb converts its respective power when plugged into constnt 120-V potentil difference. () Find the resistnce of ech bulb. (b) How long does it tke for 1.00 C to pss through the dim bulb? How is this chrge different t the time of its exit compred with the time of its entry? (c) How long does it tke for 1.00 J to pss through the dim bulb? How is this energy different t the time of its exit compred with the time of its entry? (d) Find the cost of running the dim bulb continuously for 30.0 dys if the electric compny sells its product t $ per kwh. Wht product does the electric compny sell? Wht is its price for one SI unit of this quntity? 53. A high-voltge trnsmission line with dimeter of 2.00 cm nd length of 200 km crries stedy current of A. If the conductor is copper wire with free chrge density of electrons/m 3, how long does it tke one electron to trvel the full length of the cble? 54. A high-voltge trnsmission line crries A strting t 700 kv for distnce of 100 mi. If the resistnce in the wire is /mi, wht is the power loss due to resistive losses? 55. A more generl definition of the temperture coefficient of resistivity is where is the resistivity t temperture T. () Assuming tht is constnt, show tht where 0 is the resistivity t temperture T 0. (b) Using the series expnsion ( e x 1 x for x V 1), show tht the resistivity is given pproximtely by the expression 0[1 (T T 0 )] for (T T 0 ) V A copper cble is to be designed to crry current of 300 A with power loss of only 2.00 W/m. Wht is the required rdius of the copper cble? 57. An experiment is conducted to mesure the electricl resistivity of Nichrome in the form of wires with different lengths nd cross-sectionl res. For one set of 1 d dt 0e (TT 0 ) mesurements, student uses 30-guge wire, which hs cross-sectionl re of m 2. The student mesures the potentil difference cross the wire nd the current in the wire with voltmeter nd mmeter, respectively. For ech of the mesurements given in the tble tken on wires of three different lengths, clculte the resistnce of the wires nd the corresponding vlues of the resistivity. Wht is the verge vlue of the resistivity, nd how does this vlue compre with the vlue given in Tble 27.1? L (m) V (V) I (A) R () (m) An electric utility compny supplies customer s house from the min power lines (120 V) with two copper wires, ech of which is 50.0 m long nd hs resistnce of per 300 m. () Find the voltge t the customer s house for lod current of 110 A. For this lod current, find (b) the power tht the customer is receiving nd (c) the power lost in the copper wires. 59. A stright cylindricl wire lying long the x xis hs length of m nd dimeter of mm. It is mde of mteril described by Ohm s lw with resistivity of Assume tht potentil of 4.00 V is mintined t x 0, nd tht V 0 t x m. Find () the electric field E in the wire, (b) the resistnce of the wire, (c) the electric current in the wire, nd (d) the current density J in the wire. Express vectors in vector nottion. (e) Show tht E 60. A stright cylindricl wire lying long the x xis hs length L nd dimeter d. It is mde of mteril described by Ohm s lw with resistivity. Assume tht potentil V is mintined t x 0, nd tht V 0 t x L. In terms of L, d, V,, nd physicl constnts, derive expressions for () the electric field in the wire, (b) the resistnce of the wire, (c) the electric current in the wire, nd (d) the current density in the wire. Express vectors in vector nottion. (e) Show tht E J. 61. The potentil difference cross the filment of lmp is mintined t constnt level while equilibrium temperture is being reched. It is observed tht the stedystte current in the lmp is only one tenth of the current drwn by the lmp when it is first turned on. If the temperture coefficient of resistivity for the lmp t 20.0 C is ( C) 1, nd if the resistnce increses linerly with incresing temperture, wht is the finl operting temperture of the filment? 62. The current in resistor decreses by 3.00 A when the potentil difference pplied cross the resistor decreses from 12.0 V to 6.00 V. Find the resistnce of the resistor m. J.

163 866 CHAPTER 27 Current nd Resistnce 63. An electric cr is designed to run off bnk of 12.0-V btteries with totl energy storge of J. () If the electric motor drws 8.00 kw, wht is the current delivered to the motor? (b) If the electric motor drws 8.00 kw s the cr moves t stedy speed of 20.0 m/s, how fr will the cr trvel before it is out of juice? 64. Review Problem. When stright wire is heted, its resistnce is given by the expression R R 0 [1 (T T 0 )] ccording to Eqution 27.21, where is the temperture coefficient of resistivity. () Show tht more precise result, one tht ccounts for the fct tht the length nd re of the wire chnge when heted, is where is the coefficient of liner expnsion (see Chpter 19). (b) Compre these two results for 2.00-m-long copper wire of rdius mm, first t 20.0 C nd then heted to C. 65. The temperture coefficients of resistivity in Tble 27.1 were determined t temperture of 20 C. Wht would they be t 0 C? (Hint: The temperture coefficient of resistivity t 20 C stisfies the expression 0[1 (T T 0 )], where 0 is the resistivity of the mteril t T 0 20C. The temperture coefficient of resistivity t 0 C must stisfy the expression 0 [1 T ], where 0 is the resistivity of the mteril t 0 C.) 66. A resistor is constructed by shping mteril of resistivity into hollow cylinder of length L nd with inner nd outer rdii r nd r b, respectively (Fig. P27.66). In use, the ppliction of potentil difference between the ends of the cylinder produces current prllel to the xis. () Find generl expression for the resistnce of such device in terms of L,, r, nd r b. (b) Obtin numericl vlue for R when L 4.00 cm, r cm, r nd 10 5 b 1.20 cm, m. (c) Now suppose tht the potentil difference is pplied between the inner nd outer surfces so tht the resulting current flows rdilly outwrd. Find generl expression for the resistnce of the device in terms of L,, R R 0[1 (T T 0 )][1 (T T 0 )] [1 2(T T 0 )] 3.50 r, nd r b. (d) Clculte the vlue of R, using the prmeter vlues given in prt (b). 67. In certin stereo system, ech speker hs resistnce of The system is rted t 60.0 W in ech chnnel, nd ech speker circuit includes fuse rted t 4.00 A. Is this system dequtely protected ginst overlod? Explin your resoning. 68. A close nlogy exists between the flow of energy due to temperture difference (see Section 20.7) nd the flow of electric chrge due to potentil difference. The energy dq nd the electric chrge dq re both trnsported by free electrons in the conducting mteril. Consequently, good electricl conductor is usully good therml conductor s well. Consider thin conducting slb of thickness dx, re A, nd electricl conductivity, with potentil difference dv between opposite fces. Show tht the current I dq/dt is given by the eqution on the left: Chrge conduction Anlogous therml conduction (Eq ) dq dt ka dt dx In the nlogous therml conduction eqution on the right, the rte of energy flow dq /dt (in SI units of joules per second) is due to temperture grdient dt/dx in mteril of therml conductivity k. Stte nlogous rules relting the direction of the electric current to the chnge in potentil nd relting the direction of energy flow to the chnge in temperture. 69. Mteril with uniform resistivity is formed into wedge, s shown in Figure P Show tht the resistnce between fce A nd fce B of this wedge is y 1 dq dt A dv dx R Fce A L w(y 2 y 1 ) ln y 2 y 1 Fce B y 2 r L L w r b ρ Figure P27.66 Figure P A mteril of resistivity is formed into the shpe of truncted cone of ltitude h, s shown in Figure P27.70.

164 Answers to Quick Quizzes 867 The bottom end hs rdius b, nd the top end hs rdius. Assuming tht the current is distributed uniformly over ny prticulr cross-section of the cone so tht the current density is not function of rdil position (lthough it does vry with position long the xis b Figure P27.70 h of the cone), show tht the resistnce between the two ends is given by the expression R h b 71. The currentvoltge chrcteristic curve for semiconductor diode s function of temperture T is given by the eqution I I 0 (e ev/k BT 1) Here, the first symbol e represents the bse of the nturl logrithm. The second e is the chrge on the electron. The k B is Boltzmnn s constnt, nd T is the bsolute temperture. Set up spredsheet to clculte I nd R (V )/I for V V to V in increments of V. Assume tht I na. Plot R versus V for T 280 K, 300 K, nd 320 K. ANSWERS TO QUICK QUIZZES 27.1 d, b c,. The current in prt (d) is equivlent to two positive chrges moving to the left. Prts (b) nd (c) ech represent four positive chrges moving in the sme direction becuse negtive chrges moving to the left re equivlent to positive chrges moving to the right. The current in prt () is equivlent to five positive chrges moving to the right Every portion of the wire crries the sme current even though the wire constricts. As the cross-sectionl re decreses, the drift velocity must increse in order for the constnt current to be mintined, in ccordnce with Eqution Equtions 27.5 nd 27.6 indicte tht the current density lso increses. An incresing electric field must be cusing the incresing current density, s indicted by Eqution If you were to drw this sitution, you would show the electric field lines being compressed into the smller re, indicting incresing mgnitude of the electric field The curvture of the line indictes tht the device is nonohmic (tht is, its resistnce vries with potentil difference). Being the definition of resistnce, Eqution 27.8 still pplies, giving different vlues for R t different points on the curve. The slope of the tngent to the grph line t point is the reciprocl of the dynmic resistnce t tht point. Note tht the resistnce of the device (s mesured by n ohmmeter) is the reciprocl of the slope of secnt line joining the origin to prticulr point on the curve The cble should be s short s possible but still ble to rech from one vehicle to nother (smll ), it should be quite thick (lrge A), nd it should be mde of m- teril with low resistivity. Referring to Tble 27.1, you should probbly choose copper or luminum becuse the only two mterils in the tble tht hve lower vlues silver nd gold re prohibitively expensive for your purposes Just fter it is turned on. When the filment is t room temperture, its resistnce is low, nd hence the current is reltively lrge (I V/R). As the filment wrms up, its resistnce increses, nd the current decreses. Older lightbulbs often fil just s they re turned on becuse this lrge initil current spike produces rpid temperture increse nd stress on the filment (c). Becuse the potentil difference V is the sme cross the two bulbs nd becuse the power delivered to conductor is I V, the 60-W bulb, with its higher power rting, must crry the greter current. The 30-W bulb hs the higher resistnce becuse it drws less current t the sme potentil difference I I b I c I d I e I f. The current I leves the positive terminl of the bttery nd then splits to flow through the two bulbs; thus, I I c I e. From Quick Quiz 27.6, we know tht the current in the 60-W bulb is greter thn tht in the 30-W bulb. (Note tht ll the current does not follow the pth of lest resistnce, which in this cse is through the 60-W bulb.) Becuse chrge does not build up in the bulbs, we know tht ll the chrge flowing into bulb from the left must flow out on the right; consequently, I c I d nd I e I f. The two currents leving the bulbs recombine to form the current bck into the bttery, I f I d I b.

165 P U Z Z L E R If ll these pplinces were operting t one time, circuit breker would probbly be tripped, preventing potentilly dngerous sitution. Wht cuses circuit breker to trip when too mny electricl devices re plugged into one circuit? (George Semple) c h p t e r Direct Current Circuits Chpter Outline 28.1 Electromotive Force 28.2 Resistors in Series nd in Prllel 28.3 Kirchhoff s Rules 28.4 RC Circuits 28.5 (Optionl) Electricl Instruments 28.6 (Optionl) Household Wiring nd Electricl Sfety 868

166 T his chpter is concerned with the nlysis of some simple electric circuits tht contin btteries, resistors, nd cpcitors in vrious combintions. The nlysis of these circuits is simplified by the use of two rules known s Kirchhoff s rules, which follow from the lws of conservtion of energy nd conservtion of electric chrge. Most of the circuits nlyzed re ssumed to be in stedy stte, which mens tht the currents re constnt in mgnitude nd direction. In Section 28.4 we discuss circuits in which the current vries with time. Finlly, we describe vriety of common electricl devices nd techniques for mesuring current, potentil difference, resistnce, nd emf Electromotive Force ELECTROMOTIVE FORCE In Section 27.6 we found tht constnt current cn be mintined in closed circuit through the use of source of emf, which is device (such s bttery or genertor) tht produces n electric field nd thus my cuse chrges to move round circuit. One cn think of source of emf s chrge pump. When n electric potentil difference exists between two points, the source moves chrges uphill from the lower potentil to the higher. The emf describes the work done per unit chrge, nd hence the SI unit of emf is the volt. Consider the circuit shown in Figure 28.1, consisting of bttery connected to resistor. We ssume tht the connecting wires hve no resistnce. The positive terminl of the bttery is t higher potentil thn the negtive terminl. If we neglect the internl resistnce of the bttery, the potentil difference cross it (clled the terminl voltge) equls its emf. However, becuse rel bttery lwys hs some internl resistnce r, the terminl voltge is not equl to the emf for bttery in circuit in which there is current. To understnd why this is so, consider the circuit digrm in Figure 28.2, where the bttery of Figure 28.1 is represented by the dshed rectngle contining n emf in series with n internl resistnce r. Now imgine moving through the bttery clockwise from to b nd mesuring the electric potentil t vrious loctions. As we pss from the negtive terminl to the positive terminl, the potentil increses by n mount. However, s we move through the resistnce r, the potentil decreses by n mount Ir, where I is the current in the circuit. Thus, the terminl voltge of the bttery is 1 V V b V Bttery Resistor Figure 28.1 A circuit consisting of resistor connected to the terminls of bttery. 1 The terminl voltge in this cse is less thn the emf by n mount Ir. In some situtions, the terminl voltge my exceed the emf by n mount Ir. This hppens when the direction of the current is opposite tht of the emf, s in the cse of chrging bttery with nother source of emf.

167 870 CHAPTER 28 Direct Current Circuits I ε IR V ε Figure 28.2 ε d r R () (b) () Circuit digrm of source of emf (in this cse, bttery), of internl resistnce r, connected to n externl resistor of resistnce R. (b) Grphicl representtion showing how the electric potentil chnges s the circuit in prt () is trversed clockwise. b r c c R b Ir d I V Ir (28.1) From this expression, note tht is equivlent to the open-circuit voltge tht is, the terminl voltge when the current is zero. The emf is the voltge lbeled on bttery for exmple, the emf of D cell is 1.5 V. The ctul potentil difference between the terminls of the bttery depends on the current through the bttery, s described by Eqution Figure 28.2b is grphicl representtion of the chnges in electric potentil s the circuit is trversed in the clockwise direction. By inspecting Figure 28.2, we see tht the terminl voltge V must equl the potentil difference cross the externl resistnce R, often clled the lod resistnce. The lod resistor might be simple resistive circuit element, s in Figure 28.1, or it could be the resistnce of some electricl device (such s toster, n electric heter, or lightbulb) connected to the bttery (or, in the cse of household devices, to the wll outlet). The resistor represents lod on the bttery becuse the bttery must supply energy to operte the device. The potentil difference cross the lod resistnce is V IR. Combining this expression with Eqution 28.1, we see tht IR Ir (28.2) Solving for the current gives I (28.3) R r This eqution shows tht the current in this simple circuit depends on both the lod resistnce R externl to the bttery nd the internl resistnce r. If R is much greter thn r, s it is in mny rel-world circuits, we cn neglect r. If we multiply Eqution 28.2 by the current I, we obtin I I 2 R I 2 r (28.4) This eqution indictes tht, becuse power I V (see Eq ), the totl power output I of the bttery is delivered to the externl lod resistnce in the mount I 2 R nd to the internl resistnce in the mount I 2 r. Agin, if r V R, then most of the power delivered by the bttery is trnsferred to the lod resistnce. EXAMPLE 28.1 Terminl Voltge of Bttery A bttery hs n emf of 12.0 V nd n internl resistnce of Its terminls re connected to lod resistnce of () Find the current in the circuit nd the terminl voltge of the bttery. Solution Using first Eqution 28.3 nd then Eqution 28.1, we obtin 12.0 V I 3.93 A R r 3.05 V Ir 12.0 V (3.93 A)(0.05 ) To check this result, we cn clculte the voltge cross the lod resistnce R : V IR (3.93 A)(3.00 ) 11.8 V 11.8 V (b) Clculte the power delivered to the lod resistor, the power delivered to the internl resistnce of the bttery, nd the power delivered by the bttery. Solution The power delivered to the lod resistor is R I 2 R (3.93 A) 2 (3.00 ) The power delivered to the internl resistnce is r I 2 r (3.93 A) 2 (0.05 ) 46.3 W W Hence, the power delivered by the bttery is the sum of these quntities, or 47.1 W. You should check this result, using the expression I.

168 28.2 Resistors in Series nd in Prllel 871 EXAMPLE 28.2 Mtching the Lod Show tht the mximum power delivered to the lod resistnce R in Figure 28.2 occurs when the lod resistnce mtches the internl resistnce tht is, when R r. mx Solution The power delivered to the lod resistnce is equl to I 2 R, where I is given by Eqution 28.3: I 2 R 2 R When is plotted versus R s in Figure 28.3, we find tht reches mximum vlue of 2 /4r t R r. We cn lso prove this by differentiting with respect to R, setting the result equl to zero, nd solving for R. The detils re left s problem for you to solve (Problem 57). (R r) 2 r 2r 3r Figure 28.3 Grph of the power delivered by bttery to lod resistor of resistnce R s function of R. The power delivered to the resistor is mximum when the lod resistnce equls the internl resistnce of the bttery. R 28.2 RESISTORS IN SERIES AND IN PARALLEL Suppose tht you nd your friends re t crowded bsketbll gme in sports ren nd decide to leve erly. You hve two choices: (1) your whole group cn exit through single door nd wlk down long hllwy contining severl concession stnds, ech surrounded by lrge crowd of people witing to buy food or souvenirs; or (b) ech member of your group cn exit through seprte door in the min hll of the ren, where ech will hve to push his or her wy through single group of people stnding by the door. In which scenrio will less time be required for your group to leve the ren? It should be cler tht your group will be ble to leve fster through the seprte doors thn down the hllwy where ech of you hs to push through severl groups of people. We could describe the groups of people in the hllwy s cting in series, becuse ech of you must push your wy through ll of the groups. The groups of people round the doors in the ren cn be described s cting in prllel. Ech member of your group must push through only one group of people, nd ech member pushes through different group of people. This simple nlogy will help us understnd the behvior of currents in electric circuits contining more thn one resistor. When two or more resistors re connected together s re the lightbulbs in Figure 28.4, they re sid to be in series. Figure 28.4b is the circuit digrm for the lightbulbs, which re shown s resistors, nd the bttery. In series connection, ll the chrges moving through one resistor must lso pss through the second resistor. (This is nlogous to ll members of your group pushing through the crowds in the single hllwy of the sports ren.) Otherwise, chrge would ccumulte between the resistors. Thus, for series combintion of resistors, the currents in the two resistors re the sme becuse ny chrge tht psses through R 1 must lso pss through R 2. The potentil difference pplied cross the series combintion of resistors will divide between the resistors. In Figure 28.4b, becuse the voltge drop 2 from to b 2 The term voltge drop is synonymous with decrese in electric potentil cross resistor nd is used often by individuls working with electric circuits.

169 872 CHAPTER 28 Direct Current Circuits R 1 R 2 R 1 R b 2 c R eq c I I I Bttery V V () Figure 28.4 (b) () A series connection of two resistors R 1 nd R 2. The current in R 1 is the sme s tht in R 2. (b) Circuit digrm for the two-resistor circuit. (c) The resistors replced with single resistor hving n equivlent resistnce R eq R 1 R 2. (c) equls IR 1 nd the voltge drop from b to c equls IR 2, the voltge drop from to c is V IR 1 IR 2 I(R 1 R 2 ) Therefore, we cn replce the two resistors in series with single resistor hving n equivlent resistnce R eq, where R eq R 1 R 2 (28.5) The resistnce R eq is equivlent to the series combintion R 1 R 2 in the sense tht the circuit current is unchnged when R eq replces R 1 R 2. The equivlent resistnce of three or more resistors connected in series is R eq R 1 R 2 R 3 (28.6) This reltionship indictes tht the equivlent resistnce of series connection of resistors is lwys greter thn ny individul resistnce. Quick Quiz 28.1 If piece of wire is used to connect points b nd c in Figure 28.4b, does the brightness of bulb R 1 increse, decrese, or sty the sme? Wht hppens to the brightness of bulb R 2? A series connection of three lightbulbs, ll rted t 120 V but hving power rtings of 60 W, 75 W, nd 200 W. Why re the intensities of the bulbs different? Which bulb hs the gretest resistnce? How would their reltive intensities differ if they were connected in prllel? Now consider two resistors connected in prllel, s shown in Figure When the current I reches point in Figure 28.5b, clled junction, it splits into two prts, with I 1 going through R 1 nd I 2 going through R 2. A junction is ny point in circuit where current cn split ( just s your group might split up nd leve the ren through severl doors, s described erlier.) This split results in less current in ech individul resistor thn the current leving the bttery. Becuse chrge must be conserved, the current I tht enters point must equl the totl current leving tht point: I I 1 I 2

170 28.2 Resistors in Series nd in Prllel 873 R 1 R 2 () Figure 28.5 Bttery I 1 I I 2 R 1 R 2 V (b) b I R eq V () A prllel connection of two resistors R 1 nd R 2. The potentil difference cross R 1 is the sme s tht cross R 2. (b) Circuit digrm for the two-resistor circuit. (c) The resistors replced with single resistor hving n equivlent resistnce R eq (R 1 1 R 1 2 ) 1. (c) QuickLb Tpe one pir of drinking strws end to end, nd tpe second pir side by side. Which pir is esier to blow through? Wht would hppen if you were compring three strws tped end to end with three tped side by side? As cn be seen from Figure 28.5, both resistors re connected directly cross the terminls of the bttery. Thus, when resistors re connected in prllel, the potentil differences cross them re the sme. Strws in series Strws in prllel Becuse the potentil differences cross the resistors re the sme, the expression V IR gives From this result, we see tht the equivlent resistnce of two resistors in prllel is given by or I I 1 I 2 V R 1 V V R V R 1 R 2 R eq 1 R eq 1 R 1 1 R 2 R eq 1 1 R 1 1 R 2 An extension of this nlysis to three or more resistors in prllel gives (28.7) 1 R eq 1 R 1 1 R 2 1 R 3 (28.8) The equivlent resistnce of severl resistors in prllel

171 874 CHAPTER 28 Direct Current Circuits We cn see from this expression tht the equivlent resistnce of two or more resistors connected in prllel is lwys less thn the lest resistnce in the group. Household circuits re lwys wired such tht the pplinces re connected in prllel. Ech device opertes independently of the others so tht if one is switched off, the others remin on. In ddition, the devices operte on the sme voltge. Three lightbulbs hving power rtings of 25 W, 75 W, nd 150 W, connected in prllel to voltge source of bout 100 V. All bulbs re rted t the sme voltge. Why do the intensities differ? Which bulb drws the most current? Which hs the lest resistnce? Quick Quiz 28.2 Assume tht the bttery of Figure 28.1 hs zero internl resistnce. If we dd second resistor in series with the first, does the current in the bttery increse, decrese, or sty the sme? How bout the potentil difference cross the bttery terminls? Would your nswers chnge if the second resistor were connected in prllel to the first one? Quick Quiz 28.3 Are utomobile hedlights wired in series or in prllel? How cn you tell? EXAMPLE 28.3 Find the Equivlent Resistnce Four resistors re connected s shown in Figure () Find the equivlent resistnce between points nd c. Solution The combintion of resistors cn be reduced in steps, s shown in Figure The 8.0- nd 4.0- resistors re in series; thus, the equivlent resistnce between nd b is 12 (see Eq. 28.5). The 6.0- nd 3.0- resistors re in prllel, so from Eqution 28.7 we find tht the equivlent resistnce from b to c is 2.0. Hence, the equivlent resistnce from to c is 14. (b) Wht is the current in ech resistor if potentil difference of 42V is mintined between nd c? Solution The currents in the 8.0- nd 4.0- resistors re the sme becuse they re in series. In ddition, this is the sme s the current tht would exist in the 14- equivlent resistor subject to the 42-V potentil difference. Therefore, using Eqution 27.8 (R V/I ) nd the results from prt (), we obtin I V c 42 V 3.0 A R eq 14 This is the current in the 8.0- nd 4.0- resistors. When this 3.0-A current enters the junction t b, however, it splits, with prt pssing through the 6.0- resistor (I 1 ) nd prt through the 3.0- resistor (I 2 ). Becuse the potentil difference is V bc cross ech of these resistors (since they re in prllel), we see tht (6.0 ) I 1 (3.0 )I 2, or I 2 2I 1. Using this result nd the fct tht I 1 I A, we find tht I A nd I A. We could hve guessed this t the strt by noting tht the current through the 3.0- resistor hs to be twice tht through the 6.0- resistor, in view of their reltive resistnces nd the fct tht the sme voltge is pplied to ech of them. As finl check of our results, note tht V bc (6.0 )I 1 (3.0 )I V nd V b (12 )I 36 V; therefore, V c V b V bc 42 V, s it must. () (b) (c) 8.0 Ω I 12 Ω I Ω b I 2 14 Ω Figure 28.6 b 6.0 Ω 3.0 Ω 2.0 Ω c c c

172 28.2 Resistors in Series nd in Prllel 875 EXAMPLE 28.4 Three Resistors in Prllel Three resistors re connected in prllel s shown in Figure A potentil difference of 18 V is mintined between points nd b. () Find the current in ech resistor. Solution The resistors re in prllel, nd so the potentil difference cross ech must be 18 V. Applying the reltionship V IR to ech resistor gives I 1 V R 1 I 2 V R 2 I 3 V R 3 (b) Clculte the power delivered to ech resistor nd the totl power delivered to the combintion of resistors. Solution We pply the reltionship (V ) 2 /R to ech resistor nd obtin 1 V 2 R 1 2 V 2 R 2 3 V 2 R 3 18 V V V 9.0 (18 V)2 3.0 (18 V)2 6.0 (18 V) A 3.0 A 2.0 A 110 W 54 W 36 W This shows tht the smllest resistor receives the most power. Summing the three quntities gives totl power of 200 W. (c) Clculte the equivlent resistnce of the circuit. Solution We cn use Eqution 28.8 to find R eq : Exercise R eq Use R eq to clculte the totl power delivered by the bttery. Answer 200 W. Figure R eq V I Three resistors connected in prllel. The voltge cross ech resistor is 18 V. I 1 I 2 I Ω 6.0 Ω 9.0 Ω b EXAMPLE 28.5 Finding R eq by Symmetry Arguments Consider five resistors connected s shown in Figure Find the equivlent resistnce between points nd b. Solution In this type of problem, it is convenient to ssume current entering junction nd then pply symmetry 5 Ω c 1 Ω 1 Ω 1 Ω 1 Ω 1/2 Ω 1/2 Ω 1 Ω 5 Ω b c,d b c,d b b 1 Ω 1 Ω 1 Ω 1 Ω d () (b) (c) (d) Figure 28.8 Becuse of the symmetry in this circuit, the 5- resistor does not contribute to the resistnce between points nd b nd therefore cn be disregrded when we clculte the equivlent resistnce.

173 876 CHAPTER 28 Direct Current Circuits rguments. Becuse of the symmetry in the circuit (ll 1- resistors in the outside loop), the currents in brnches c nd d must be equl; hence, the electric potentils t points c nd d must be equl. This mens tht V cd 0 nd, s result, points c nd d my be connected together without ffecting the circuit, s in Figure 28.8b. Thus, the 5- resistor my be removed from the circuit nd the remining circuit then reduced s in Figures 28.8c nd d. From this reduction we see tht the equivlent resistnce of the combintion is 1. Note tht the result is 1 regrdless of the vlue of the resistor connected between c nd d. CONCEPTUAL EXAMPLE 28.6 Figure 28.9 illustrtes how three-wy lightbulb is constructed to provide three levels of light intensity. The socket of the lmp is equipped with three-wy switch for selecting different light intensities. The bulb contins two filments. When the lmp is connected to 120-V source, one filment receives 100 W of power, nd the other receives 75 W. Explin how the two filments re used to provide three different light intensities. Solution The three light intensities re mde possible by pplying the 120 V to one filment lone, to the other filment lone, or to the two filments in prllel. When switch S 1 is closed nd switch S 2 is opened, current psses only through the 75-W filment. When switch S 1 is open nd switch S 2 is closed, current psses only through the 100-W filment. When both switches re closed, current psses through both filments, nd the totl power is 175 W. If the filments were connected in series nd one of them were to brek, no current could pss through the bulb, nd the bulb would give no illumintion, regrdless of the switch position. However, with the filments connected in prllel, if one of them (for exmple, the 75-W filment) breks, the bulb will still operte in two of the switch positions s current psses through the other (100-W) filment. Opertion of Three-Wy Lightbulb Exercise Determine the resistnces of the two filments nd their prllel equivlent resistnce. Answer 144, 192, Figure W filment 75-W filment S 1 S V A three-wy lightbulb. APPLICATION Strings of Lights Strings of lights re used for mny ornmentl purposes, such s decorting Christms trees. Over the yers, both prllel nd series connections hve been used for multilight strings powered by 120 V. 3 Series-wired bulbs re sfer thn prllel-wired bulbs for indoor Christms-tree use becuse series-wired bulbs operte with less light per bulb nd t lower temperture. However, if the filment of single bulb fils (or if the bulb is removed from its socket), ll the lights on the string re extinguished. The populrity of series-wired light strings diminished becuse troubleshooting filed bulb ws tedious, time-consuming chore tht involved trilnd-error substitution of good bulb in ech socket long the string until the defective bulb ws found. In prllel-wired string, ech bulb opertes t 120 V. By design, the bulbs re brighter nd hotter thn those on series-wired string. As result, these bulbs re inherently more dngerous (more likely to strt fire, for instnce), but if one bulb in prllel-wired string fils or is removed, the rest of the bulbs continue to glow. (A 25-bulb string of 4-W bulbs results in power of 100 W; the totl power becomes substntil when severl strings re used.) A new design ws developed for so-clled miniture lights wired in series, to prevent the filure of one bulb from extinguishing the entire string. The solution is to crete connection (clled jumper) cross the filment fter it fils. (If n lternte connection existed cross the filment before 3 These nd other household devices, such s the three-wy lightbulb in Conceptul Exmple 28.6 nd the kitchen pplinces shown in this chpter s Puzzler, ctully operte on lternting current (c), to be introduced in Chpter 33.

174 28.3 Kirchhoff s Rules 877 it filed, ech bulb would represent prllel circuit; in this circuit, the current would flow through the lternte connection, forming short circuit, nd the bulb would not glow.) When the filment breks in one of these miniture lightbulbs, 120 V ppers cross the bulb becuse no current is present in the bulb nd therefore no drop in potentil occurs cross the other bulbs. Inside the lightbulb, smll loop covered by n insulting mteril is wrpped round the filment leds. An rc burns the insultion nd connects the filment leds when 120 V ppers cross the bulb tht is, when the filment fils. This short now completes the circuit through the bulb even though the filment is no longer ctive (Fig ). Suppose tht ll the bulbs in 50-bulb miniture-light string re operting. A 2.4-V potentil drop occurs cross ech bulb becuse the bulbs re in series. The power input to this style of bulb is 0.34 W, so the totl power supplied to the string is only 17 W. We clculte the filment resistnce t the operting temperture to be (2.4 V) 2 /(0.34 W) 17. When the bulb fils, the resistnce cross its terminls is reduced to zero becuse of the lternte jumper connection mentioned in the preceding prgrph. All the other bulbs not only sty on but glow more brightly becuse the totl resistnce of the string is reduced nd consequently the current in ech bulb increses. Let us ssume tht the operting resistnce of bulb remins t 17 even though its temperture rises s result of the incresed current. If one bulb fils, the potentil drop cross ech of the remining bulbs increses to 2.45 V, the current increses from 0.142A to A, nd the power increses to W. As more lights fil, the current keeps rising, the filment of ech bulb opertes t higher temperture, nd the lifetime of the bulb is reduced. It is therefore good ide to check for filed (nonglowing) bulbs in such series-wired string nd replce them s soon s possible, in order to mximize the lifetimes of ll the bulbs. Filment Jumper Glss insultor () (b) Figure () Schemtic digrm of modern miniture holidy lightbulb, with jumper connection to provide current pth if the filment breks. (b) A Christms-tree lightbulb KIRCHHOFF S RULES As we sw in the preceding section, we cn nlyze simple circuits using the expression V IR nd the rules for series nd prllel combintions of resistors. Very often, however, it is not possible to reduce circuit to single loop. The procedure for nlyzing more complex circuits is gretly simplified if we use two principles clled Kirchhoff s rules: 1. The sum of the currents entering ny junction in circuit must equl the sum of the currents leving tht junction: I in I out (28.9)

175 878 CHAPTER 28 Direct Current Circuits 2. The sum of the potentil differences cross ll elements round ny closed circuit loop must be zero: V 0 closed loop (28.10) Kirchhoff s first rule is sttement of conservtion of electric chrge. All current tht enters given point in circuit must leve tht point becuse chrge cnnot build up t point. If we pply this rule to the junction shown in Figure 28.11, we obtin Gustv Kirchhoff ( ) Kirchhoff, professor t Heidelberg, Germny, nd Robert Bunsen invented the spectroscope nd founded the science of spectroscopy, which we shll study in Chpter 40. They discovered the elements cesium nd rubidium nd invented stronomicl spectroscopy. Kirchhoff formulted nother Kirchhoff s rule, nmely, cool substnce will bsorb light of the sme wvelengths tht it emits when hot. (AIP ESVA/W. F. Meggers Collection) QuickLb Drw n rbitrrily shped closed loop tht does not cross over itself. Lbel five points on the loop, b, c, d, nd e, nd ssign rndom number to ech point. Now strt t nd work your wy round the loop, clculting the difference between ech pir of djcent numbers. Some of these differences will be positive, nd some will be negtive. Add the differences together, mking sure you ccurtely keep trck of the lgebric signs. Wht is the sum of the differences ll the wy round the loop? I 1 I 2 I 3 Figure 28.11b represents mechnicl nlog of this sitution, in which wter flows through brnched pipe hving no leks. The flow rte into the pipe equls the totl flow rte out of the two brnches on the right. Kirchhoff s second rule follows from the lw of conservtion of energy. Let us imgine moving chrge round the loop. When the chrge returns to the strting point, the chrge circuit system must hve the sme energy s when the chrge strted from it. The sum of the increses in energy in some circuit elements must equl the sum of the decreses in energy in other elements. The potentil energy decreses whenever the chrge moves through potentil drop IR cross resistor or whenever it moves in the reverse direction through source of emf. The potentil energy increses whenever the chrge psses through bttery from the negtive terminl to the positive terminl. Kirchhoff s second rule pplies only for circuits in which n electric potentil is defined t ech point; this criterion my not be stisfied if chnging electromgnetic fields re present, s we shll see in Chpter 31. In justifying our clim tht Kirchhoff s second rule is sttement of conservtion of energy, we imgined crrying chrge round loop. When pplying this rule, we imgine trveling round the loop nd consider chnges in electric potentil, rther thn the chnges in potentil energy described in the previous prgrph. You should note the following sign conventions when using the second rule: Becuse chrges move from the high-potentil end of resistor to the lowpotentil end, if resistor is trversed in the direction of the current, the chnge in potentil V cross the resistor is IR (Fig ). If resistor is trversed in the direction opposite the current, the chnge in potentil V cross the resistor is IR (Fig b). If source of emf (ssumed to hve zero internl resistnce) is trversed in the direction of the emf (from to ), the chnge in potentil V is (Fig c). The emf of the bttery increses the electric potentil s we move through it in this direction. If source of emf (ssumed to hve zero internl resistnce) is trversed in the direction opposite the emf (from to ), the chnge in potentil V is (Fig d). In this cse the emf of the bttery reduces the electric potentil s we move through it. Limittions exist on the numbers of times you cn usefully pply Kirchhoff s rules in nlyzing given circuit. You cn use the junction rule s often s you need, so long s ech time you write n eqution you include in it current tht hs not been used in preceding junction-rule eqution. In generl, the number of times you cn use the junction rule is one fewer thn the number of junction

176 28.3 Kirchhoff s Rules 879 I 1 I 2 Flow in Figure I 3 () (b) Flow out () Kirchhoff s junction rule. Conservtion of chrge requires tht ll current entering junction must leve tht junction. Therefore, I 1 I 2 I 3. (b) A mechnicl nlog of the junction rule: the mount of wter flowing out of the brnches on the right must equl the mount flowing into the single brnch on the left. () (b) (c) I V = IR I V = IR ε V = ε ε ε (d) b V = εε Figure Rules for determining the potentil chnges cross resistor nd bttery. (The bttery is ssumed to hve no internl resistnce.) Ech circuit element is trversed from left to right. b b b points in the circuit. You cn pply the loop rule s often s needed, so long s new circuit element (resistor or bttery) or new current ppers in ech new eqution. In generl, in order to solve prticulr circuit problem, the number of independent equtions you need to obtin from the two rules equls the number of unknown currents. Complex networks contining mny loops nd junctions generte gret numbers of independent liner equtions nd correspondingly gret number of unknowns. Such situtions cn be hndled formlly through the use of mtrix lgebr. Computer progrms cn lso be written to solve for the unknowns. The following exmples illustrte how to use Kirchhoff s rules. In ll cses, it is ssumed tht the circuits hve reched stedy-stte conditions tht is, the currents in the vrious brnches re constnt. Any cpcitor cts s n open circuit; tht is, the current in the brnch contining the cpcitor is zero under stedystte conditions. Problem-Solving Hints Kirchhoff s Rules Drw circuit digrm, nd lbel ll the known nd unknown quntities. You must ssign direction to the current in ech brnch of the circuit. Do not be lrmed if you guess the direction of current incorrectly; your result will be negtive, but its mgnitude will be correct. Although the ssignment of current directions is rbitrry, you must dhere rigorously to the ssigned directions when pplying Kirchhoff s rules. Apply the junction rule to ny junctions in the circuit tht provide new reltionships mong the vrious currents.

177 880 CHAPTER 28 Direct Current Circuits Apply the loop rule to s mny loops in the circuit s re needed to solve for the unknowns. To pply this rule, you must correctly identify the chnge in potentil s you imgine crossing ech element in trversing the closed loop (either clockwise or counterclockwise). Wtch out for errors in sign! Solve the equtions simultneously for the unknown quntities. EXAMPLE 28.7 A Single-Loop Circuit A single-loop circuit contins two resistors nd two btteries, s shown in Figure (Neglect the internl resistnces of the btteries.) () Find the current in the circuit. Solution We do not need Kirchhoff s rules to nlyze this simple circuit, but let us use them nywy just to see how they re pplied. There re no junctions in this single-loop circuit; thus, the current is the sme in ll elements. Let us ssume tht the current is clockwise, s shown in Figure Trversing the circuit in the clockwise direction, strting t, we see tht : b represents potentil chnge of 1, b : c represents potentil chnge of IR 1, c : d represents potentil chnge of 2, nd d : represents potentil chnge of IR 2. Applying Kirchhoff s loop rule gives Figure d ε 1 = 6.0 V R 2 = 10 Ω I R 1 = 8.0 Ω 2 = 12 V A series circuit contining two btteries nd two resistors, where the polrities of the btteries re in opposition. ε b c Solving for I nd using the vlues given in Figure 28.13, we obtin The negtive sign for I indictes tht the direction of the current is opposite the ssumed direction. (b) Wht power is delivered to ech resistor? Wht power is delivered by the 12-V bttery? Solution I 1 2 R 1 R 2 V 0 1 IR 1 2 IR V 12 V I 2 R 1 (0.33 A) 2 (8.0 ) 2 I 2 R 2 (0.33 A) 2 (10 ) 0.33 A 0.87 W 1.1 W Hence, the totl power delivered to the resistors is W. The 12-V bttery delivers power I W. Hlf of this power is delivered to the two resistors, s we just clculted. The other hlf is delivered to the 6-V bttery, which is being chrged by the 12-V bttery. If we hd included the internl resistnces of the btteries in our nlysis, some of the power would pper s internl energy in the btteries; s result, we would hve found tht less power ws being delivered to the 6-V bttery. EXAMPLE 28.8 Applying Kirchhoff s Rules Find the currents I 1, I 2, nd I 3 in the circuit shown in Figure Solution Notice tht we cnnot reduce this circuit to simpler form by mens of the rules of dding resistnces in series nd in prllel. We must use Kirchhoff s rules to nlyze this circuit. We rbitrrily choose the directions of the currents s lbeled in Figure Applying Kirchhoff s junction rule to junction c gives (1) I 1 I 2 I 3 We now hve one eqution with three unknowns I 1, I 2, nd I 3. There re three loops in the circuit bcd, befcb, nd efd. We therefore need only two loop equtions to determine the unknown currents. (The third loop eqution would give no new informtion.) Applying Kirchhoff s loop rule to loops bcd nd befcb nd trversing these loops clockwise, we obtin the expressions (2) bcd 10 V (6 )I 1 (2 )I 3 0 (3) befcb 14 V (6 )I 1 10 V (4 )I 2 0

178 28.3 Kirchhoff s Rules 881 Note tht in loop befcb we obtin positive vlue when trversing the 6- resistor becuse our direction of trvel is opposite the ssumed direction of I 1. Expressions (1), (2), nd (3) represent three independent equtions with three unknowns. Substituting Eqution (1) into Eqution (2) gives 10 V (6 )I 1 (2 ) (I 1 I 2 ) 0 (4) 10 V (8 )I 1 (2 )I 2 Dividing ech term in Eqution (3) by 2 nd rerrnging gives (5) 12 V (3 )I 1 (2 )I 2 Subtrcting Eqution (5) from Eqution (4) elimintes I 2, giving Using this vlue of I 1 in Eqution (5) gives vlue for I 2 : I 2 3 A 22 V (11 )I 1 I 1 2 A (2 )I 2 (3 )I 1 12 V (3 ) (2 A) 12 V 6 V e 14 V f Finlly, I 3 I 1 I 2 1 A b 4 Ω 10 V 6 Ω I 1 c I 2 I 3 The fct tht I 2 nd I 3 re both negtive indictes only tht the currents re opposite the direction we chose for them. However, the numericl vlues re correct. Wht would hve hppened hd we left the current directions s lbeled in Figure but trversed the loops in the opposite direction? Figure d 2 Ω A circuit contining three loops. Exercise nd c. Answer 2 V. Find the potentil difference between points b EXAMPLE 28.9 A Multiloop Circuit () Under stedy-stte conditions, find the unknown currents I 1, I 2, nd I 3 in the multiloop circuit shown in Figure Solution First note tht becuse the cpcitor represents n open circuit, there is no current between g nd b long pth ghb under stedy-stte conditions. Therefore, when the chrges ssocited with I 1 rech point g, they ll go through the 8.00-V bttery to point b; hence, I gb I 1. Lbeling the currents s shown in Figure nd pplying Eqution 28.9 to junction c, we obtin (1) I 1 I 2 I 3 Eqution pplied to loops defcd nd cfgbc, trversed clockwise, gives (2) defcd 4.00 V (3.00 )I 2 (5.00 )I 3 0 (3) cfgbc (3.00 )I 2 (5.00 )I V 0 From Eqution (1) we see tht I 1 I 3 I 2, which, when substituted into Eqution (3), gives (4) (8.00 )I 2 (5.00 )I V 0 Subtrcting Eqution (4) from Eqution (2), we eliminte I 3 nd find tht I V A Becuse our vlue for I 2 is negtive, we conclude tht the direction of I 2 is from c to f through the resistor. Despite Figure I 3 I 1 d c b 5.00 Ω I V 4.00 V e 3.00 Ω 8.00 V 5.00 Ω I = µ F A multiloop circuit. Kirchhoff s loop rule cn be pplied to ny closed loop, including the one contining the cpcitor. f g h I 3 I 1

179 882 CHAPTER 28 Direct Current Circuits this interprettion of the direction, however, we must continue to use this negtive vlue for I 2 in subsequent clcultions becuse our equtions were estblished with our originl choice of direction. Using I A in Equtions (3) nd (1) gives (b) Wht is the chrge on the cpcitor? Solution I A I A We cn pply Kirchhoff s loop rule to loop bghb (or ny other loop tht contins the cpcitor) to find the potentil difference V cp cross the cpcitor. We enter this potentil difference in the eqution without reference to sign convention becuse the chrge on the cpcitor depends only on the mgnitude of the potentil difference. Moving clockwise round this loop, we obtin 8.00 V V cp 3.00 V 0 V cp 11.0 V Becuse Q C V cp (see Eq. 26.1), the chrge on the cpcitor is Q (6.00 F)(11.0 V) 66.0 C Why is the left side of the cpcitor positively chrged? Exercise Find the voltge cross the cpcitor by trversing ny other loop. Answer 11.0 V. Exercise Reverse the direction of the 3.00-V bttery nd nswer prts () nd (b) gin. Answer () (b) 30 C. I A, I A, I A; 28.4 RC CIRCUITS So fr we hve been nlyzing stedy-stte circuits, in which the current is constnt. In circuits contining cpcitors, the current my vry in time. A circuit contining series combintion of resistor nd cpcitor is clled n RC circuit. Chrging Cpcitor Let us ssume tht the cpcitor in Figure is initilly unchrged. There is no current while switch S is open (Fig b). If the switch is closed t t 0, however, chrge begins to flow, setting up current in the circuit, nd the cpcitor begins to chrge. 4 Note tht during chrging, chrges do not jump cross the cpcitor pltes becuse the gp between the pltes represents n open circuit. Insted, chrge is trnsferred between ech plte nd its connecting wire due to the electric field estblished in the wires by the bttery, until the cpcitor is fully chrged. As the pltes become chrged, the potentil difference cross the cpcitor increses. The vlue of the mximum chrge depends on the voltge of the bttery. Once the mximum chrge is reched, the current in the circuit is zero becuse the potentil difference cross the cpcitor mtches tht supplied by the bttery. To nlyze this circuit quntittively, let us pply Kirchhoff s loop rule to the circuit fter the switch is closed. Trversing the loop clockwise gives q C IR 0 (28.11) where q/c is the potentil difference cross the cpcitor nd IR is the potentil 4 In previous discussions of cpcitors, we ssumed stedy-stte sitution, in which no current ws present in ny brnch of the circuit contining cpcitor. Now we re considering the cse before the stedy-stte condition is relized; in this sitution, chrges re moving nd current exists in the wires connected to the cpcitor.

180 28.4 RC Circuits 883 Resistor Cpcitor R R Switch C q q I Figure Bttery () ε S (b) t < 0 (c) t > 0 () A cpcitor in series with resistor, switch, nd bttery. (b) Circuit digrm representing this system t time t 0, before the switch is closed. (c) Circuit digrm t time t 0, fter the switch hs been closed. ε S difference cross the resistor. We hve used the sign conventions discussed erlier for the signs on nd IR. For the cpcitor, notice tht we re trveling in the direction from the positive plte to the negtive plte; this represents decrese in potentil. Thus, we use negtive sign for this voltge in Eqution Note tht q nd I re instntneous vlues tht depend on time (s opposed to stedy-stte vlues) s the cpcitor is being chrged. We cn use Eqution to find the initil current in the circuit nd the mximum chrge on the cpcitor. At the instnt the switch is closed (t 0), the chrge on the cpcitor is zero, nd from Eqution we find tht the initil current in the circuit I 0 is mximum nd is equl to I 0 R (current t t 0) (28.12) Mximum current At this time, the potentil difference from the bttery terminls ppers entirely cross the resistor. Lter, when the cpcitor is chrged to its mximum vlue Q, chrges cese to flow, the current in the circuit is zero, nd the potentil difference from the bttery terminls ppers entirely cross the cpcitor. Substituting I 0 into Eqution gives the chrge on the cpcitor t this time: Q C (mximum chrge) (28.13) To determine nlyticl expressions for the time dependence of the chrge nd current, we must solve Eqution single eqution contining two vribles, q nd I. The current in ll prts of the series circuit must be the sme. Thus, the current in the resistnce R must be the sme s the current flowing out of nd into the cpcitor pltes. This current is equl to the time rte of chnge of the chrge on the cpcitor pltes. Thus, we substitute I dq /dt into Eqution nd rerrnge the eqution: dq dt R To find n expression for q, we first combine the terms on the right-hnd side: dq C dt RC q C q RC RC q RC Mximum chrge on the cpcitor

181 884 CHAPTER 28 Direct Current Circuits Now we multiply by dt nd divide by q C to obtin Integrting this expression, using the fct tht q 0 t t 0, we obtin q 0 dq q C 1 RC dt dq q C 1 t RC ln q C C t RC From the definition of the nturl logrithm, we cn write this expression s 0 dt Chrge versus time for cpcitor being chrged q(t ) C (1 e t/rc ) Q(1 e t /RC ) (28.14) where e is the bse of the nturl logrithm nd we hve mde the substitution C Q from Eqution We cn find n expression for the chrging current by differentiting Eqution with respect to time. Using I dq /dt, we find tht Current versus time for chrging cpcitor I(t ) R e t /RC (28.15) Plots of cpcitor chrge nd circuit current versus time re shown in Figure Note tht the chrge is zero t t 0 nd pproches the mximum vlue C s t :. The current hs its mximum vlue I 0 /R t t 0 nd decys exponentilly to zero s t :. The quntity RC, which ppers in the exponents of Equtions nd 28.15, is clled the time constnt of the circuit. It represents the time it tkes the current to decrese to 1/e of its initil vlue; tht is, in time, I e 1 I In time 2, I e I 0. I I 0, nd so forth. Likewise, in time, the chrge increses from zero to C (1 e 1 ) 0.632C. The following dimensionl nlysis shows tht hs the units of time: [] [RC] V Q I V Q [t] T Q /t q I C ε Cε τ =RC I 0 ε I 0 = R 0.368I 0 τ t τ t Figure () (b) () Plot of cpcitor chrge versus time for the circuit shown in Figure After time intervl equl to one time constnt hs pssed, the chrge is 63.2% of the mximum vlue C. The chrge pproches its mximum vlue s t pproches infinity. (b) Plot of current versus time for the circuit shown in Figure The current hs its mximum vlue I 0 /R t t 0 nd decys to zero exponentilly s t pproches infinity. After time intervl equl to one time constnt hs pssed, the current is 36.8% of its initil vlue.

182 28.4 RC Circuits 885 RC Becuse hs units of time, the combintion t /RC is dimensionless, s it must be in order to be n exponent of e in Equtions nd Q The energy output of the bttery s the cpcitor is fully chrged is After the cpcitor is fully chrged, the energy stored in the cpcitor 1 is Q C C2, which is just hlf the energy output of the bttery. It is left s problem (Problem 60) to show tht the remining hlf of the energy supplied by the bttery ppers s internl energy in the resistor. Dischrging Cpcitor Now let us consider the circuit shown in Figure 28.18, which consists of cpcitor crrying n initil chrge Q, resistor, nd switch. The initil chrge Q is not the sme s the mximum chrge Q in the previous discussion, unless the dischrge occurs fter the cpcitor is fully chrged (s described erlier). When the switch is open, potentil difference Q /C exists cross the cpcitor nd there is zero potentil difference cross the resistor becuse I 0. If the switch is closed t t 0, the cpcitor begins to dischrge through the resistor. At some time t during the dischrge, the current in the circuit is I nd the chrge on the cpcitor is q (Fig b). The circuit in Figure is the sme s the circuit in Figure except for the bsence of the bttery. Thus, we eliminte the emf from Eqution to obtin the pproprite loop eqution for the circuit in Figure 28.18: q C IR 0 (28.16) C C Q Q q q S t < 0 () R R I When we substitute I dq /dt into this expression, it becomes R dq dt Integrting this expression, using the fct tht q Q t t 0, gives q Q q C dq q 1 RC dt dq q 1 RC t ln q Q t RC q(t ) Qe t /RC 0 dt (28.17) Figure S t > 0 (b) () A chrged cpcitor connected to resistor nd switch, which is open t t 0. (b) After the switch is closed, current tht decreses in mgnitude with time is set up in the direction shown, nd the chrge on the cpcitor decreses exponentilly with time. Chrge versus time for dischrging cpcitor Differentiting this expression with respect to time gives the instntneous current s function of time: I(t) dq dt d dt (Qe t /RC ) Q RC e t /RC (28.18) Current versus time for dischrging cpcitor where Q /RC I 0 is the initil current. The negtive sign indictes tht the current direction now tht the cpcitor is dischrging is opposite the current direction when the cpcitor ws being chrged. (Compre the current directions in Figs c nd 28.18b.) We see tht both the chrge on the cpcitor nd the current decy exponentilly t rte chrcterized by the time constnt RC.

183 886 CHAPTER 28 Direct Current Circuits CONCEPTUAL EXAMPLE Mny utomobiles re equipped with windshield wipers tht cn operte intermittently during light rinfll. How does the opertion of such wipers depend on the chrging nd dischrging of cpcitor? Solution The wipers re prt of n RC circuit whose time constnt cn be vried by selecting different vlues of R Intermittent Windshield Wipers through multiposition switch. As it increses with time, the voltge cross the cpcitor reches point t which it triggers the wipers nd dischrges, redy to begin nother chrging cycle. The time intervl between the individul sweeps of the wipers is determined by the vlue of the time constnt. EXAMPLE Chrging Cpcitor in n RC Circuit An unchrged cpcitor nd resistor re connected in series to bttery, s shown in Figure If 12.0 V, C 5.00 F, nd R , find the time constnt of the circuit, the mximum chrge on the cpcitor, the mximum current in the circuit, nd the chrge nd current s functions of time. Solution The time constnt of the circuit is ( )( F) 4.00 s. The mximum chrge on the cpcitor is Q C (5.00 F) (12.0 V) 60.0 C. The mximum current in the circuit is I 0 /R (12.0 V)/( ) 15.0 A. Using these vlues nd Equtions nd 28.15, we find tht q(t) I(t) (60.0 C)(1 e t/4.00 s ) (15.0 A) e t/4.00 s Grphs of these functions re provided in Figure RC Exercise Clculte the chrge on the cpcitor nd the current in the circuit fter one time constnt hs elpsed. Answer 37.9 C, 5.52 A. q(µc) µ Q = 60.0 µc µ I(µA) µ () t = τ I 0 = 15.0 µa µ t(s) R 5 t = τ C ε S Figure The switch of this series RC circuit, open for times t 0, is closed t t 0. Figure (b) t(s) Plots of () chrge versus time nd (b) current versus time for the RC circuit shown in Figure 28.19, with 12.0 V, R , nd C 5.00 F. EXAMPLE Dischrging Cpcitor in n RC Circuit Consider cpcitor of cpcitnce C tht is being dischrged through resistor of resistnce R, s shown in Figure () After how mny time constnts is the chrge on the cpcitor one-fourth its initil vlue? Solution The chrge on the cpcitor vries with time ccording to Eqution 28.17, q(t) Qe t /RC. To find the time it tkes q to drop to one-fourth its initil vlue, we substitute q(t) Q /4 into this expression nd solve for t:

184 28.5 Electricl Instruments 887 Tking logrithms of both sides, we find ln 4 Q 4 t RC Qe t /RC 1 4 e t /RC t RC(ln 4) 1.39RC 1.39 (b) The energy stored in the cpcitor decreses with time s the cpcitor dischrges. After how mny time constnts is this stored energy one-fourth its initil vlue? Solution Using Equtions (U Q 2 /2C) nd 28.17, we cn express the energy stored in the cpcitor t ny time t s where U 0 Q 2 /2C is the initil energy stored in the cpcitor. As in prt (), we now set U U 0 /4 nd solve for t: Agin, tking logrithms of both sides nd solving for t gives Exercise After how mny time constnts is the current in the circuit one-hlf its initil vlue? Answer U q 2 2C (Qet /RC ) 2 2C t 1 2RC(ln 4) 0.693RC 0.693RC Q 2 2C e 2t /RC U 0 e 2t /RC U 0 4 U 2t /RC 0e 1 4 e 2t /RC EXAMPLE Energy Delivered to Resistor A 5.00-F cpcitor is chrged to potentil difference of 800 V nd then dischrged through 25.0-k resistor. How much energy is delivered to the resistor in the time it tkes to fully dischrge the cpcitor? Solution We shll solve this problem in two wys. The first wy is to note tht the initil energy in the circuit equls the energy stored in the cpcitor, C 2 /2 (see Eq ). Once the cpcitor is fully dischrged, the energy stored in it is zero. Becuse energy is conserved, the initil energy stored in the cpcitor is trnsformed into internl energy in the resistor. Using the given vlues of C nd, we find Energy 1 2 C2 1 2 ( F)(800 V) J The second wy, which is more difficult but perhps more instructive, is to note tht s the cpcitor dischrges through the resistor, the rte t which energy is delivered to the resistor is given by I 2 R, where I is the instntneous current given by Eqution Becuse power is defined s the time rte of chnge of energy, we conclude tht the energy delivered to the resistor must equl the time integrl of I 2 R dt: To evlute this integrl, we note tht the initil current is equl to /R nd tht ll prmeters except t re constnt. Thus, we find (1) Energy 2 e 2t/RC dt R 0 This integrl hs vlue of RC/2; hence, we find which grees with the result we obtined using the simpler pproch, s it must. Note tht we cn use this second pproch to find the totl energy delivered to the resistor t ny time fter the switch is closed by simply replcing the upper limit in the integrl with tht specific vlue of t. Exercise Energy I 2 R dt (I 0 e t /RC ) 2 R dt 0 Energy 1 2 C2 Show tht the integrl in Eqution (1) hs the vlue RC/2. 0 I 0 Optionl Section 28.5 The Ammeter ELECTRICAL INSTRUMENTS A device tht mesures current is clled n mmeter. The current to be mesured must pss directly through the mmeter, so the mmeter must be connected in se-

185 888 CHAPTER 28 Direct Current Circuits A Figure R 1 ε R 2 Current cn be mesured with n mmeter connected in series with the resistor nd bttery of circuit. An idel mmeter hs zero resistnce. R 1 Figure ε V R 2 The potentil difference cross resistor cn be mesured with voltmeter connected in prllel with the resistor. An idel voltmeter hs infinite resistnce. ries with other elements in the circuit, s shown in Figure When using n mmeter to mesure direct currents, you must be sure to connect it so tht current enters the instrument t the positive terminl nd exits t the negtive terminl. Idelly, n mmeter should hve zero resistnce so tht the current being mesured is not ltered. In the circuit shown in Figure 28.21, this condition requires tht the resistnce of the mmeter be much less thn R 1 R 2. Becuse ny mmeter lwys hs some internl resistnce, the presence of the mmeter in the circuit slightly reduces the current from the vlue it would hve in the meter s bsence. The Voltmeter A device tht mesures potentil difference is clled voltmeter. The potentil difference between ny two points in circuit cn be mesured by ttching the terminls of the voltmeter between these points without breking the circuit, s shown in Figure The potentil difference cross resistor R 2 is mesured by connecting the voltmeter in prllel with R 2. Agin, it is necessry to observe the polrity of the instrument. The positive terminl of the voltmeter must be connected to the end of the resistor tht is t the higher potentil, nd the negtive terminl to the end of the resistor t the lower potentil. An idel voltmeter hs infinite resistnce so tht no current psses through it. In Figure 28.22, this condition requires tht the voltmeter hve resistnce much greter thn R 2. In prctice, if this condition is not met, corrections should be mde for the known resistnce of the voltmeter. The Glvnometer The glvnometer is the min component in nlog mmeters nd voltmeters. Figure illustrtes the essentil fetures of common type clled the D Arsonvl glvnometer. It consists of coil of wire mounted so tht it is free to rotte on pivot in mgnetic field provided by permnent mgnet. The bsic op- Scle N S Figure Spring () Coil () The principl components of D Arsonvl glvnometer. When the coil situted in mgnetic field crries current, the mgnetic torque cuses the coil to twist. The ngle through which the coil rottes is proportionl to the current in the coil becuse of the countercting torque of the spring. (b) A lrge-scle model of glvnometer movement. Why does the coil rotte bout the verticl xis fter the switch is closed? (b)

186 28.5 Electricl Instruments 889 Glvnometer 60 Ω Glvnometer R s 60 Ω R p Figure () () When glvnometer is to be used s n mmeter, shunt resistor R p is connected in prllel with the glvnometer. (b) When the glvnometer is used s voltmeter, resistor R s is connected in series with the glvnometer. (b) ertion of the glvnometer mkes use of the fct tht torque cts on current loop in the presence of mgnetic field (Chpter 29). The torque experienced by the coil is proportionl to the current through it: the lrger the current, the greter the torque nd the more the coil rottes before the spring tightens enough to stop the rottion. Hence, the deflection of needle ttched to the coil is proportionl to the current. Once the instrument is properly clibrted, it cn be used in conjunction with other circuit elements to mesure either currents or potentil differences. A typicl off-the-shelf glvnometer is often not suitble for use s n mmeter, primrily becuse it hs resistnce of bout 60. An mmeter resistnce this gret considerbly lters the current in circuit. You cn understnd this by considering the following exmple: The current in simple series circuit contining 3-V bttery nd 3- resistor is 1 A. If you insert 60- glvnometer in this circuit to mesure the current, the totl resistnce becomes 63 nd the current is reduced to A! A second fctor tht limits the use of glvnometer s n mmeter is the fct tht typicl glvnometer gives full-scle deflection for currents of the order of 1 ma or less. Consequently, such glvnometer cnnot be used directly to mesure currents greter thn this vlue. However, it cn be converted to useful mmeter by plcing shunt resistor R p in prllel with the glvnometer, s shown in Figure The vlue of R p must be much less thn the glvnometer resistnce so tht most of the current to be mesured psses through the shunt resistor. A glvnometer cn lso be used s voltmeter by dding n externl resistor R s in series with it, s shown in Figure 28.24b. In this cse, the externl resistor must hve vlue much greter thn the resistnce of the glvnometer to ensure tht the glvnometer does not significntly lter the voltge being mesured. The Whetstone Bridge An unknown resistnce vlue cn be ccurtely mesured using circuit known s Whetstone bridge (Fig ). This circuit consists of the unknown resistnce R x, three known resistnces R 1, R 2, nd R 3 (where R 1 is clibrted vrible resistor), glvnometer, nd bttery. The known resistor R 1 is vried until the glvnometer reding is zero tht is, until there is no current from to b. Under this condition the bridge is sid to be blnced. Becuse the electric potentil t Figure I 1 I 2 R 1 R 2 R 3 Circuit digrm for Whetstone bridge, n instrument used to mesure n unknown resistnce R x in terms of known resistnces R 1, R 2, nd R 3. When the bridge is blnced, no current is present in the glvnometer. The rrow superimposed on the circuit symbol for resistor R 1 indictes tht the vlue of this resistor cn be vried by the person operting the bridge. G R x b

187 890 CHAPTER 28 Direct Current Circuits The strin guge, device used for experimentl stress nlysis, consists of thin coiled wire bonded to flexible plstic bcking. The guge mesures stresses by detecting chnges in the resistnce of the coil s the strip bends. Resistnce mesurements re mde with this device s one element of Whetstone bridge. Strin guges re commonly used in modern electronic blnces to mesure the msses of objects. Figure Voltges, currents, nd resistnces re frequently mesured with digitl multimeters like this one. point must equl the potentil t point b when the bridge is blnced, the potentil difference cross R 1 must equl the potentil difference cross R 2. Likewise, the potentil difference cross R 3 must equl the potentil difference cross R x. From these considertions we see tht (1) I 1 R 1 I 2 R 2 (2) I 1 R 3 I 2 R x Dividing Eqution (1) by Eqution (2) elimintes the currents, nd solving for R x, we find tht R x R 2R 3 (28.19) R 1 A number of similr devices lso operte on the principle of null mesurement (tht is, djustment of one circuit element to mke the glvnometer red zero). One exmple is the cpcitnce bridge used to mesure unknown cpcitnces. These devices do not require clibrted meters nd cn be used with ny voltge source. Whetstone bridges re not useful for resistnces bove 10 5, but modern electronic instruments cn mesure resistnces s high s Such instruments hve n extremely high resistnce between their input terminls. For exmple, input resistnces of re common in most digitl multimeters, which re devices tht re used to mesure voltge, current, nd resistnce (Fig ). The Potentiometer A potentiometer is circuit tht is used to mesure n unknown emf x by comprison with known emf. In Figure 28.27, point d represents sliding contct tht is used to vry the resistnce (nd hence the potentil difference) between points nd d. The other required components re glvnometer, bttery of known emf 0, nd bttery of unknown emf x. With the currents in the directions shown in Figure 28.27, we see from Kirchhoff s junction rule tht the current in the resistor R x is I I x, where I is the current in the left brnch (through the bttery of emf 0 ) nd I x is the current in the right brnch. Kirchhoff s loop rule pplied to loop bcd trversed clockwise gives x (I I x )R x 0 Becuse current I x psses through it, the glvnometer displys nonzero reding. The sliding contct t d is now djusted until the glvnometer reds zero (indicting blnced circuit nd tht the potentiometer is nother null-mesurement device). Under this condition, the current in the glvnometer is zero, nd the potentil difference between nd d must equl the unknown emf x : x IR x Next, the bttery of unknown emf is replced by stndrd bttery of known emf s, nd the procedure is repeted. If R s is the resistnce between nd d when blnce is chieved this time, then s IR s where it is ssumed tht I remins the sme. Combining this expression with the preceding one, we see tht x R x (28.20) R s s

188 28.6 Household Wiring nd Electricl Sfety 891 If the resistor is wire of resistivity, its resistnce cn be vried by using the sliding contct to vry the length L, indicting how much of the wire is prt of the circuit. With the substitutions R s L s /A nd R x L x /A, Eqution becomes x L x L s (28.21) where L x is the resistor length when the bttery of unknown emf x is in the circuit nd L s is the resistor length when the stndrd bttery is in the circuit. The sliding-wire circuit of Figure without the unknown emf nd the glvnometer is sometimes clled voltge divider. This circuit mkes it possible to tp into ny desired smller portion of the emf 0 by djusting the length of the resistor. s ε ε 0 I I I x Figure d R x Circuit digrm for potentiometer. The circuit is used to mesure n unknown emf x. I x b G c ε x Optionl Section 28.6 HOUSEHOLD WIRING AND ELECTRICAL SAFETY Household circuits represent prcticl ppliction of some of the ides presented in this chpter. In our world of electricl pplinces, it is useful to understnd the power requirements nd limittions of conventionl electricl systems nd the sfety mesures tht prevent ccidents. In conventionl instlltion, the utility compny distributes electric power to individul homes by mens of pir of wires, with ech home connected in prllel to these wires. One wire is clled the live wire, 5 s illustrted in Figure 28.28, nd the other is clled the neutrl wire. The potentil difference between these two wires is bout 120 V. This voltge lterntes in time, with the neutrl wire connected to ground nd the potentil of the live wire oscillting reltive to ground. Much of wht we hve lerned so fr for the constnt-emf sitution (direct current) cn lso be pplied to the lternting current tht power compnies supply to businesses nd households. (Alternting voltge nd current re discussed in Chpter 33.) A meter is connected in series with the live wire entering the house to record the household s usge of electricity. After the meter, the wire splits so tht there re severl seprte circuits in prllel distributed throughout the house. Ech circuit contins circuit breker (or, in older instlltions, fuse). The wire nd circuit breker for ech circuit re crefully selected to meet the current demnds for tht circuit. If circuit is to crry currents s lrge s 30 A, hevy wire nd n pproprite circuit breker must be selected to hndle this current. A circuit used to power only lmps nd smll pplinces often requires only 15 A. Ech circuit hs its own circuit breker to ccommodte vrious lod conditions. As n exmple, consider circuit in which toster oven, microwve oven, nd coffee mker re connected (corresponding to R 1, R 2, nd R 3 in Figure nd s shown in the chpter-opening photogrph). We cn clculte the current drwn by ech pplince by using the expression I V. The toster oven, rted t W, drws current of W/120 V 8.33 A. The microwve oven, rted t W, drws 10.8 A, nd the coffee mker, rted t 800 W, drws 6.67 A. If the three pplinces re operted simultneously, they drw totl cur- 5 Live wire is common expression for conductor whose electric potentil is bove or below ground potentil. Live Neutrl R 1 Figure Meter Circuit breker R 2 R 3 Wiring digrm for household circuit. The resistnces represent pplinces or other electricl devices tht operte with n pplied voltge of 120 V. 120 V 0 V

189 892 CHAPTER 28 Direct Current Circuits Figure A power connection for 240-V pplince. rent of 25.8 A. Therefore, the circuit should be wired to hndle t lest this much current. If the rting of the circuit breker protecting the circuit is too smll sy, 20 A the breker will be tripped when the third pplince is turned on, preventing ll three pplinces from operting. To void this sitution, the toster oven nd coffee mker cn be operted on one 20-A circuit nd the microwve oven on seprte 20-A circuit. Mny hevy-duty pplinces, such s electric rnges nd clothes dryers, require 240 V for their opertion (Fig ). The power compny supplies this voltge by providing third wire tht is 120 V below ground potentil. The potentil difference between this live wire nd the other live wire (which is 120 V bove ground potentil) is 240 V. An pplince tht opertes from 240-V line requires hlf the current of one operting from 120-V line; therefore, smller wires cn be used in the higher-voltge circuit without overheting. Figure A three-pronged power cord for 120-V pplince. Electricl Sfety When the live wire of n electricl outlet is connected directly to ground, the circuit is completed nd short-circuit condition exists. A short circuit occurs when lmost zero resistnce exists between two points t different potentils; this results in very lrge current. When this hppens ccidentlly, properly operting circuit breker opens the circuit nd no dmge is done. However, person in contct with ground cn be electrocuted by touching the live wire of fryed cord or other exposed conductor. An exceptionlly good (lthough very dngerous) ground contct is mde when the person either touches wter pipe (normlly t ground potentil) or stnds on the ground with wet feet. The ltter sitution represents good ground becuse norml, nondistilled wter is conductor becuse it contins lrge number of ions ssocited with impurities. This sitution should be voided t ll cost. Electric shock cn result in ftl burns, or it cn cuse the muscles of vitl orgns, such s the hert, to mlfunction. The degree of dmge to the body depends on the mgnitude of the current, the length of time it cts, the prt of the body touched by the live wire, nd the prt of the body through which the current psses. Currents of 5 ma or less cuse senstion of shock but ordinrily do little or no dmge. If the current is lrger thn bout 10 ma, the muscles contrct nd the person my be unble to relese the live wire. If current of bout 100 ma psses through the body for only few seconds, the result cn be ftl. Such lrge current prlyzes the respirtory muscles nd prevents brething. In some cses, currents of bout 1 A through the body cn produce serious (nd sometimes ftl) burns. In prctice, no contct with live wires is regrded s sfe whenever the voltge is greter thn 24 V. Mny 120-V outlets re designed to ccept three-pronged power cord such s the one shown in Figure (This feture is required in ll new electricl instlltions.) One of these prongs is the live wire t nominl potentil of 120 V. The second, clled the neutrl, is nominlly t 0 V nd crries current to ground. The third, round prong is sfety ground wire tht normlly crries no current but is both grounded nd connected directly to the csing of the pplince. If the live wire is ccidentlly shorted to the csing (which cn occur if the wire insultion wers off), most of the current tkes the low-resistnce pth through the pplince to ground. In contrst, if the csing of the pplince is not properly grounded nd short occurs, nyone in contct with the pplince experiences n electric shock becuse the body provides low-resistnce pth to ground.